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aboutthesettle.com | 1,656,813,253,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104209449.64/warc/CC-MAIN-20220703013155-20220703043155-00774.warc.gz | 121,906,541 | 40,952 | Asked By: George Cox Date: created: Sep 28 2021
## What is God’s number
Answered By: Howard Mitchell Date: created: Sep 30 2021
The term “God’s number” is sometimes given to the graph diameter of Rubik’s graph, which is the minimum number of turns required to solve a Rubik’s cube from an arbitrary starting position (i.e., in the worst case).
Rokicki et al..
Asked By: Herbert Harris Date: created: Jul 23 2021
## What is special about the number 39
Answered By: George Gonzales Date: created: Jul 25 2021
39 is the sum of consecutive prime numbers. 39 is the sum of the first 3 powers of 3. 39 is the smallest natural number which has three partitions into three parts. … 39 is the actual number of lashes given by the Sanhedrin to a person meted the punishment of 40 lashes.
Asked By: Reginald King Date: created: Jan 10 2022
## Why is 39 an evil number
Answered By: Cody Thomas Date: created: Jan 10 2022
An NPR report explains: “Many Afghans say that the number 39 translates into morda-gow, which literally means ‘dead cow’ but is also a well-known slang term for a procurer of prostitutes—a pimp.” So when Afghans see a car with number 39 on the license plate, they head the other way.
Asked By: Aidan Ward Date: created: Oct 14 2021
## What is the meaning of 1
Answered By: Miguel Sanchez Date: created: Oct 14 2021
unit, unityWikipedia. 1. 1 (one, also called unit, unity, and (multiplicative) identity) is a number, and a numerical digit used to represent that number in numerals. It represents a single entity, the unit of counting or measurement.
Asked By: Julian Thomas Date: created: Mar 16 2022
## What is the definition of symbolism
Answered By: John Diaz Date: created: Mar 17 2022
Symbolism is a literary device that uses symbols, be they words, people, marks, locations, or abstract ideas to represent something beyond the literal meaning. The concept of symbolism is not confined to works of literature: symbols inhabit every corner of our daily life.
Asked By: Timothy Hernandez Date: created: Feb 01 2021
## What is Jesus’s number
Answered By: Gabriel Diaz Date: created: Feb 02 2021
888In Christian numerology, the number 888 represents Jesus, or sometimes more specifically Christ the Redeemer. This representation may be justified either through gematria, by counting the letter values of the Greek transliteration of Jesus’ name, or as an opposing value to 666, the number of the beast.
Asked By: Douglas Gonzales Date: created: Mar 10 2022
## What does the number 39 mean in numerology
Answered By: Evan Butler Date: created: Mar 10 2022
The numerology number 39 is a creative number with an inclination toward assisting humankind. … The number 39 is social, optimistic, tolerant, and inspiring. It supports and encourages the creative expression of the talents inherent in others.
Asked By: Gordon Mitchell Date: created: Nov 27 2021
## What does numbers in the Bible mean
Answered By: Kevin Williams Date: created: Nov 28 2021
Numbers are a Biblical means God speaks to His people.
Asked By: Gerld Wood Date: created: Feb 08 2022
## What is an angel number
Answered By: Leonars Sanchez Date: created: Feb 11 2022
Angel numbers are a part of numerology. People who believe in angel numbers, believe that angel numbers explain why each number is connected to a frequency other than its numeric value. When believers keep seeing the same angel numbers again, they believe that guardian angels are showing them supernatural signs.
Asked By: Abraham Moore Date: created: May 27 2022
## What is the most lucky number
Answered By: Martin Bailey Date: created: May 29 2022
SevenSeven was the most popular choice for both men and women. The survey revealed some other findings, too. The distribution of favourite numbers was less for women, 85% of whom chose a number less than 30. By contrast, 85% of men chose a number lower than 64.
Asked By: Mason Ward Date: created: Jan 02 2022
## Is 39 a bad number
Answered By: Alejandro Kelly Date: created: Jan 03 2022
In Afghanistan, being called a pimp is offensive, and calling someone a pimp could carry deadly consequences. Similarly, being associated with the number 39 — whether it’s on a vehicle license plate, an apartment number or a post office box — is considered a great shame.
Asked By: Harry Price Date: created: Jun 18 2021
## Is 8 an evil number
Answered By: Jordan Garcia Date: created: Jun 20 2021
In the United States, octophobia is not that common because the number eight is not a ubiquitous symbol of bad luck, the devil or other superstitions. Americans are more likely to have a phobia of the numbers 13 (triskaidekaphobia) and 666 (hexakosioihexekontahexaphobia) because of their negative connotations.
Asked By: Timothy Peterson Date: created: Mar 31 2022
## What is God’s real name
Answered By: Gordon Barnes Date: created: Mar 31 2022
Yahweh… Hebrew personal name for God, YHWH (commonly transcribed “Yahweh”), is predominantly used, those in……
Asked By: Colin Clark Date: created: Jan 29 2022
## What does 39 mean
Answered By: Dominic Lopez Date: created: Jan 29 2022
Thank You39 means “Thank You”.
Asked By: Alfred Rodriguez Date: created: Mar 17 2021
## What does 39 mean in love
Answered By: Isaac Bailey Date: created: Mar 19 2021
The meaning of 39 when it comes to Love. The angel number 39 wants to reassure you that when it comes to love, there will always be emotional hiccups and difficulties. Remember that everything happens for a reason, and that the challenges you are facing now will soon be replaced with something wonderful.
Asked By: Cole Perez Date: created: Jun 20 2021
## What does the number 40 mean
Answered By: Jake Walker Date: created: Jun 23 2021
Spiritually, angel number 40 is a sign of encouragement from the alpha and omega and a divine message sent to humans by the guardian angels themselves as a confirmation that they’re on the right track with whatever goals they’re about to attain.
Asked By: Stanley Parker Date: created: Jul 18 2021
## What number means death
Answered By: Henry Long Date: created: Jul 20 2021
Unlucky Numbers: 四, the number 4: As you said, this number is associated with death, 死 (sǐ).
Professional
### Is It OK To Have A Chocolate Bar A Day?
Is a chocolate bar a day bad for you? If you eat a chocolate bunny a day, there is an obvious risk of becoming a major chubbo.However, a little chocolate has health benefits.The risk increased with added chocolate.Other studies have found that moderate amounts of chocolate seem to lower blood pressure. Will eating one chocolate bar a day make me fat? One chocolate bar certainly won't derail any diet, but regularly adding chocolate to your diet might tip the scales if you're not including it in your daily calorie count. And while dark chocolate has some serious health benefits, you can gain weight from eating too much of that as well. But the news is not all bad. Is it OK to eat a chocolate bar once a week? Eating Chocolate At Least Once A Week Can Change Your Brain. Good news for your sweet tooth! According to a recent…
Professional
### Question: Where Is The Spear Of Destiny Right Now?
What is a real name of Jesus? Jesus' name in Hebrew was “Yeshua” which translates to English as Joshua..... Who invented the sign of the cross? The use of the sign of the cross traces back to early Christianity, with the second century Apostolic Tradition directing that it be used during the minor exorcism of baptism, during ablutions before praying at fixed prayer times, and in times of temptation. What did the centurion say at the cross? The modern World English Bible translates the passage as: Now the centurion, and those who were with him watching Jesus, when they saw the earthquake, and the things that were done, feared exceedingly, saying, "Truly this was the Son of God." What were the names of the two criminals crucified with Jesus? In apocryphal writings, the impenitent thief is given the name Gestas, which first appears in the Gospel of Nicodemus, while his…
Professional
### Quick Answer: Why Do I Crave Salt During Period?
A woman may experience a variety of physical and emotional changes in the days leading up to her menstrual period.These changes are known as premenstrual syndrome (PMS).Food cravings, including a craving for salty foods, are a common symptom.These cravings may be related to hormonal fluctuations. What deficiency causes salt cravings? Your adrenal glands are responsible for producing hormones that are vital to your survival. Addison's disease is a rare disease that can decrease the amount of hormones produced by your adrenal glands. People with this disease experience salt cravings, in addition to other symptoms: severe fatigue or lack of energy. Is it normal to have cravings during period? As every person is different, the types of symptoms and cravings women experience before and during their period vary a great deal. One reason for appetite changes and food cravings is a change in hormones. Carbohydrates are known to promote serotonin release,…
Professional
### What Happens If We Release Sperm Daily?
Masturbating daily can lead to weakness, fatigue, early ejaculation and may inhibit sexual activities with your partner.On the other hand, missing out on regular orgasms increases stress levels and can add to mental health issues, frustration, and unhappiness in general. How many times should a man release sperm in a week? According to ATTN, it found that those who blew their load “five to seven times per week were 36 percent less likely to develop prostate cancer compared to those who ejaculated less than two times per week.” Is it good to take out sperm everyday? Men undergoing IVF treatment are routinely told to stop having sex to avoid a drop in sperm count. The new research suggests, however, that rather than reducing the chances of conception, daily sex can increase it. Daily sex also seemed to make the sperm more active, or motile, which is known to improve fertility.…
Professional
### Question: What Can I Eat Instead Of Chocolate?
Here are 19 foods that can help you fight your sugar cravings. Fruit. Share on Pinterest. Berries. Berries are an excellent, nutritious choice for stopping sugar cravings. Dark Chocolate. Snack Bars. Chia Seeds. Sugar-Free Chewing Gum or Mints. Legumes. Yogurt. What can I eat in place of chocolate? This article details 18 healthy foods that can satisfy your urge to eat without sabotaging your diet ( 2 ). Fresh Fruit. Fruit is naturally very sweet and a great choice when you get a sugar craving. Greek Yogurt. A Hot Drink. Snack Bar. Dark Chocolate. Fruit and Nut Butter. Cottage Cheese. Banana Ice Cream. What to eat if you are craving chocolate? Fill up on healthy fats like olive oil, nuts, and avocados. Eat a well-balanced diet that incorporates lots of lean protein, fruits, vegetables, and whole grains. Eat organic nut butters with no added sugar. Satisfy your sweet tooth with…
Guest
### Question: How Many Times Does A Man Get Erect In A Day?
The average man has about 11 erections each day and several more at night.But these erections don't always happen because a man is sexually excited.Sometimes, there is no sexual stimulus at all.For example, many men wake up with an erection. How long can the average man stay erect? According to a 2005 Journal of Sexual Medicine study of 500 couples across Europe and the US, the average erection during sex lasts 5.4 minutes. But “normal” erections can last anywhere from a few minutes to hours or even longer. How many times does a man erect at night? Yes, men get erections while they sleep. In fact, on average, most guys get five erections every single night. What age does a man stop getting hard? Most older men suffer not ED but erection dissatisfaction. Starting around age 50 (often earlier among smokers and, or, diabetics), erections change. In some men, the…
Guest
### What Foods Increase Female Pheromones?
What is a natural pheromone? Your natural pheromones, which are produced by the body, might have been at play.These chemicals are believed to create a natural scent that attracts individuals of the opposite sex to one another. Can you be attracted to someone's pheromones? Pheromones can also have the opposite effect and make you sexually repellent to some. People who have a love at first sight reaction to someone or who feel a strong attraction to another person are usually experiencing a pheromone attraction. How are pheromones produced? Production of Pheromones Pheromones are secreted by a variety of specialized glands in insects and mammals. Pheromones (or signature odors) are produced by bacterial and fungal fermentation of fatty acids and other compounds secreted in the anal glands of dogs, badgers, and other carnivores. How do pheromones affect behavior? Hormones usually work internally, and they only have a direct effect on the…
Guest
### How Can I Last Longer In Bed Raw?
13 Ways to Last Longer in Bed Sex Positions Can Make You Last Longer. Reduce performance anxiety. Practice Premature Ejaculation Exercises. Masturbate More Often or Have More Sex. Use Desensitizing Sprays, Creams or Gels. Improve Your Sexual Communication Skills. Take Your Time. Help Her to Orgasm First. How can a man last longer in bed naturally? Here are five methods to try. Take it slow. To get your guy to last longer, have him start slow, Men's Fitness magazine suggests. Do it again and again. A round of vigorous foreplay before sex can work wonders, according to Cosmopolitan magazine. Try pelvic floor exercises. Use a condom. Switch up the position. How do I last longer before coming? If this is the case, here are five things you can do to help him last longer in bed.5 tips to help stop premature ejaculation before it ruins your sex life Let him…
Guest
### Which Dark Chocolate Is Best For Brain?
Cocoa flavanols can improve mental function and benefit brain activity.Flavanols can help reduce blood pressure.Eating dark chocolate is good for your memory, blood pressure and your mood. Which chocolate is best for brain? The good news isn't over yet. Dark chocolate may also improve the function of your brain. One study of healthy volunteers showed that eating high-flavanol cocoa for five days improved blood flow to the brain ( 20 ). Cocoa may also significantly improve cognitive function in elderly people with mental impairment. Which Dark chocolate is best? We Tried 5 Dark Chocolate Bars, and This Is the Best One How We Graded Them.From Worst… To Best.Hershey's Special Dark.Ghirardelli Intense Dark 60% Cacao Evening Dream.Cadbury Royal Dark.Lindt Excellence 70% Cocoa Smooth Dark.Godiva 72% Cacao Dark Chocolate. Does dark chocolate help your memory? Dark Chocolate Improves Memory, Reduces Stress. Researchers say eating dark chocolate can change your brain wave frequency,…
Guest
### What Deficiency Causes Chocolate Cravings?
Cravings for specific foods could be an indication of a deficiency in a micro or macro nutrient.In particular, a craving for chocolate could highlight a magnesium deficiency.Magnesium is an essential mineral and is required for over 300 enzyme reactions in the body. What is craving chocolate a sign of? In fact, intense food cravings can be a sign you're deficient in certain nutrients. What it means: Chocolate is rich in magnesium, so strong cravings could indicate a deficiency in a mineral vital for your skin and hair. Why am I craving chocolate all of a sudden? Chocolate is the most craved food One of the most common examples I have found is the craving for chocolate. Supposedly, a craving for chocolate is a sign that your body is low on magnesium (see the table below). As it turns out, raw cacao is one of the most naturally magnesium-rich foods there…
Professor
### Quick Answer: Can Chocolate Make You Angry?
Eating chocolate can produce rage, paranoia and anger.For a small percentage of the population, eating chocolate can produce rage, paranoia and anger that occur without warning.Fortunately, for most of us, this is not the typical reaction to eating chocolate. Does chocolate cause mood swings? Can food affect your mood? Chocolate is a known mood booster, as cocoa raises serotonin levels in the brain. Can chocolate make you moody? Chocolate is reported to cause headache, obesity, rectal itching, heart burn and emotional problems like irritability, confusion, anger and depression. Chocolate is rich in carbohydrates, which increase the rate with which tryptophan enters the brain. What foods make you angry? Here are a few foods that may cause anger, anxiety, or spaciness, according to Eastern Medicine experts. Tomatoes Can Cause Anger. Eggplants Can Cause Anger. Greasy Foods Can Cause Anger. Cauliflower & Broccoli Can Cause Anxiety. Dried Fruits & Chips Can Cause…
Professor
### Does Chocolate Help With Mood Swings?
These feel-good foods can help to improve your mood and reduce feelings of anxiety.Chocolate: Dark chocolate is my favourite and most effective mood lifter.Enjoying a square or two of dark chocolate can stimulate the production of endorphins, the chemical in brain which triggers the feeling of pleasure and euphoria.18 Apr 2018 Is chocolate good for mental health? So, the bottom line here is that eating dark chocolate is good for your memory, blood pressure, and your mood. It helps alleviate depression and also acts as an anti-inflammatory, which means that it is good for your brain. And if it is good for your brainit is good for you.31 Oct 2014 Why does chocolate make you feel better? Extra reasons why chocolate makes you happy Especially dark chocolate stimulates the production of endorphins in our brain. This feeling is passed on to our brain and triggers to production of endorphins. Besides…
Professor
### Question: What Can I Substitute For Chocolate Cravings?
Following are seven smart alternatives to help you conquer your cocoa cravings—and avoid succumbing to the post-chocolate sugar slump: Reach for your favorite fruit. Go for the yogurt. Nut butter to the rescue. Hit the trail (mix) Get loco with cocoa (powder, that is) Consider carob. Wait it out. What is a good substitute for chocolate? 7 Chocolate Alternatives to Satisfy Your Sweet Tooth Carob. Carob is finely ground powder which has a similar taste to cocoa powder. Cocoa powder. Cocoa powder can curb chocolate cravings without the excess calories of chocolate. Chocolate-dipped strawberries. Frozen chocolate banana. Dark chocolate. Fruits. Cocoa nibs. What can I eat instead of chocolate? Here are 19 foods that can help you fight your sugar cravings. Fruit. Share on Pinterest. Berries. Berries are an excellent, nutritious choice for stopping sugar cravings. Dark Chocolate. Snack Bars. Chia Seeds. Sugar-Free Chewing Gum or Mints. Legumes. Yogurt. What…
Professor
### What Happens When You Have Too Many Endorphins?
One type of endorphins has twenty times the pain-blocking ability of morphine.Too many endorphins, for too long a time, can put people on edge, triggering the fight-or-flight reflex for any small event.If the body is flooded with endorphins, it naturally assumes that something painful is coming. How does endorphins affect the body? When you exercise, your body releases chemicals called endorphins. These endorphins interact with the receptors in your brain that reduce your perception of pain. Endorphins also trigger a positive feeling in the body, similar to that of morphine. What happens when you dont have enough endorphins? In general, if your body isn't producing enough endorphins, you might experience: depression. anxiety. moodiness. What activities increase endorphins? The following are seven endorphin-boosting activities to turn to in times of trouble, or when you just need a little lift. Exercise. Eat chocolate and chili peppers. Drink wine. Have sex. Get a…
Professor
### Question: What Hormones Are Released After Eating?
THURSDAY, Aug.31, 2017 (HealthDay News) -- Eating prompts the brain to release "feel good" hormones, known as endorphins, a new study shows.Researchers found the regulation of these naturally occurring opioids, which can produce a sense of pleasure or euphoria, may help the body know when it's satisfied. What hormones increase after eating? What is ghrelin? Ghrelin is a hormone that is produced and released mainly by the stomach with small amounts also released by the small intestine, pancreas and brain. Ghrelin has numerous functions. It is termed the 'hunger hormone' because it stimulates appetite, increases food intake and promotes fat storage. What chemical is released when you eat food? One of these brain chemicals is dopamine, which is released when people or animals eat tasty foods. What happens after we eat? Once proteins, fats and carbohydrates are digested, absorption takes place in the small intestine. Most of the digestion occurs…
User
### Question: Is It OK To Eat Banana Before Bed?
Do Eat: Banana Bananas are mostly made up of fast-digesting carbs.And fast digestion is definitely your goal when you're snacking before bed, Morse says.“Bananas are also a good source of magnesium, which helps calm stress hormones and so can promote sleep,” Morse says.Murray agrees that bananas are a safe pick. What happens if we eat banana before sleep? Eating a banana before bed will increase your intake in nutrients that optimize a good night's sleep. Mineral deficiency in magnesium, potassium, vitamins, as well as an amino acid called tryptophan, can cause sleeping problems. Can we take banana at night? Bananas. Best: Bananas help promote sleep because they contain the natural muscle-relaxants magnesium and potassium, says Gans. They're also carbs which will help make you sleepy as well. Can we eat banana at night for weight loss? The Morning Banana Diet will probably work simply because you'll eat less. There is…
User
### Quick Answer: Can You Eat Chocolate When Sick?
Cocoa contains a chemical called theobromine that can help the body fight off the symptoms of the common cold.Blocking the sensory nerves will stop the cough reflex in a common cold.Researchers of this study found the chocolate chemical to be more effective than codeine when treating a chronic cough. Can you eat chocolate when you have the flu? Chocolate for Cold and Flu Symptoms? While you're running to the store for tissues and cough drops, you might want to throw in some dark chocolate. That's right: Research shows that dark chocolate containing at least 70% cacao may fight cold and flu symptoms. Can you eat chocolate when you have a sore throat? That's why sugary and sticky substances like honey can help when you're suffering from a cough. So even though it's chocolate combined with medicine that can help your sore throat, not the food itself, Morice added that, "slowly…
User
### Quick Answer: Does Dark Chocolate Contain Serotonin?
Dark chocolate may increase serotonin levels not only due to the serotonin and L-tryptophan it contains, but also because it contains carbohydrates in the form of sugar, which can signal the body to produce more serotonin. Does chocolate affect serotonin levels? Because it can increase serotonin levels in the brain, dark chocolate also may increase serotonin production in the gut, and thus help your immune system. Is dark chocolate good for brain? Could Improve Brain Function. The good news isn't over yet. Dark chocolate may also improve the function of your brain. One study of healthy volunteers showed that eating high-flavanol cocoa for five days improved blood flow to the brain ( 20 ). Does dark chocolate have flavonoids? Flavonoids are a large group of plant-based antioxidants. The type of flavonoids found in chocolate are flavanols. Dark chocolate contains 45 to 80 percent cocoa solids, and milk chocolate has 5…
User
### Question: Does Dark Chocolate Contain Estrogen?
Is dark chocolate good for hormones? Dark chocolate helps to lower stress hormones, such as cortisol and catecholamines.Energy levels are boosted, and anxiety levels may be beneficially affected by chocolate.By affecting serotonin levels, chocolate may help keep depression at bay during the dark winter months. What foods are high in estrogen? Estrogen Rich Foods You Should Include in Your Diet Flax Seeds. Flax seeds are one of the richest sources of estrogen, and they top the list of foods containing phytoestrogen. Soy. Soy products have isoflavones that are known to boost the estrogen levels in women. Fruits. Nuts. Dry Fruits. Red Wine. On the other hand. How can I raise my estrogen levels quickly? There are 4 ways to naturally increase your progesterone levels: Eat foods that support your bodies production of hormones.Avoid foods and external substances that can knock your hormones out of sync.Reduce the amount of stress in… | 5,496 | 24,842 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2022-27 | longest | en | 0.926834 |
https://learn.digilentinc.com/Documents/309 | 1,537,315,958,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155792.23/warc/CC-MAIN-20180918225124-20180919005124-00359.warc.gz | 559,400,233 | 6,548 | # Zeros and Ones
## Binary System and Binary Signal
A signal in a digital circuit is a circuit net that transports an output voltage from one device to one or more input connections of other devices. In a digital circuit, signals are constrained to be at one of two voltages, either Vdd or GND. Thus, all data in digital circuits are represented by signals that can be in one of only two states, and all data operations combine two-state data inputs to produce two-state data outputs. Systems that use two-state data are known as binary systems, and a two-state signal is a binary signal. The set of voltage values {Vdd, GND} that define the state of a signal wire in a digital system are commonly represented by the numeric symbols {1, 0}, with '1' representing Vdd and '0' representing GND. Since digital systems can only represent two-state data, and since we have already assigned those states the numeric symbols '0' and '1', it follows that data in digital symbols can be represented by binary (base two) numbers. One signal wire in a digital circuit can carry one binary digit (abbreviated to “bit”) of information; groupings of signal wires can carry multiple bits that can define a binary number. Using bits to represent data in digital systems makes is easy to adopt existing logical and numerical techniques to the study of digital circuits. For example, an AND relationship can be logically described as “true” when all inputs are “true” (i.e., output Y is “true” when inputs “A”, “B”, and “C” are all “true”). If we assign the symbol '1' to “true”, then the AND relationship yields a '1' when the inputs are all '1', concisely demonstrated by the truth table in Fig. 1. Since a '1' represents Vdd and a '0' represents GND, this logical AND truth table can define a logic circuit that outputs a '1' (or Vdd) whenever all inputs are a '1'.
A group of individual digital signals may be thought of as a logical group of signals that define a multi-bit data element. Such a logical grouping of signals is called a bus. Because each signal on a bus can carry a '1' or a '0', busses can carry binary numbers. For example, if a 4-bit bus is used to represent a 4-bit binary number, then the bus can carry a binary number from 0 to 15 (0000 to 1111).
## Representing an Analog Signal in a Digital System with Zeros and Ones
In contrast to digital circuits, analog circuits use signals whose voltage levels are not constrained to two distinct levels, but instead can assume any value between Vdd and GND. Many input devices, particularly those using electronic sensors (e.g., microphones, cameras, thermometers, pressure sensors, motion and proximity detectors, etc.) produce analog voltages at their outputs. In modern electronic devices, it is likely that such signals will be converted to digital signals before they are used within the device. For example, a digital voice-memo recording device uses an analog microphone circuit to convert sound pressure waves into voltage waves on an internal circuit node. A special circuit called an analog-to-digital converter, or ADC, converts that analog voltage to a binary number that can be represented as a bus in a digital circuit. An ADC functions by taking samples of the input analog signal, measuring the magnitude of the input voltage signal (usually with reference to GND), and assigning a binary number to the measured magnitude. Once an analog signal has been converted to a binary number, a bus can carry that digital information around a circuit. In a similar manner, digital signals can be reconstituted into analog signals using a digital-to-analog converter. Thus, a binary number that represents a sample of an audio waveform can be converted to an analog signal that can, for example, drive a speaker.
## Noise of a Digital Signal
Analog signals are sensitive to noise sources and loss of signal strength over time and distance, but digital signals are relatively insensitive to noise and loss of strength. This is because a digital signal has two wide voltage bands that define a '0' and '1', and any voltage inside a band is an acceptable encoding. In Fig. 2, a digital signal with tens or hundreds of millivolts of noise defines stable 0's and 1's despite the noise; if this same amount of signal noise were present on an analog signal, the circuit could not possibly function correctly. It is because digital signals are more robust and easier to work with that electronic industries the world over have “gone digital”.
## Important Ideas
• Systems that use two-state data are known as Binary systems, and the signals that represent those two-states of data are called Binary signals.
• One signal in a circuit can carry one binary number; this signal is called a bit.
• Many electronic devices still use analog circuits that are then converted in an analog-to-digital converter (ADC) and then the digital signal is used within the device.
• Compared with Analog signals, Digital signals are relatively insensitive to noise and loss of strength. | 1,066 | 5,025 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-39 | latest | en | 0.914294 |
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A096970 Number of ways to number the cells of an n X n square grid with 1,2,3,...,n^2 so that successive integers are in the same row or column. 9
1, 8, 1512, 22394880, 50657369241600, 28606505102329400524800, 5959275438217048853558620520448000 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 COMMENTS Suggested by Leroy Quet, Jul 05 2004. For n >= 2, number of (directed) Hamiltonian paths on the n X n rook graph. - Eric W. Weisstein, Dec 16 2013 LINKS Eric Weisstein's World of Mathematics, Hamiltonian Path Eric Weisstein's World of Mathematics, Rook Graph EXAMPLE Among the 4 X 4 grids counted is: 1 2 3 10 15 6 5 11 14 13 4 12 16 7 8 9 CROSSREFS Cf. A096969, A269561, A269565. Sequence in context: A172938 A252176 A252763 * A248386 A114617 A162014 Adjacent sequences: A096967 A096968 A096969 * A096971 A096972 A096973 KEYWORD nonn,more,hard AUTHOR John W. Layman, Jul 16 2004 EXTENSIONS a(5) from Eric W. Weisstein, Dec 28 2013 a(6)-a(7) from Andrew Howroyd, Feb 29 2016 STATUS approved
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Last modified March 28 20:56 EDT 2023. Contains 361596 sequences. (Running on oeis4.) | 482 | 1,496 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2023-14 | latest | en | 0.667547 |
http://forums.na.leagueoflegends.com/board/showthread.php?s=&t=2617012&goto=nextnewest | 1,432,341,920,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207926924.76/warc/CC-MAIN-20150521113206-00076-ip-10-180-206-219.ec2.internal.warc.gz | 96,737,678 | 9,480 | ### Welcome to the Forum Archive!
Years of conversation fill a ton of digital pages, and we've kept all of it accessible to browse or copy over. Whether you're looking for reveal articles for older champions, or the first time that Rammus rolled into an "OK" thread, or anything in between, you can find it here. When you're finished, check out the boards to join in the latest League of Legends discussions.
### Noob Question Regarding AS Vayne
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Simiun
Junior Member
Hey guys. I have been playing LoL for a month or so and had an attack speed question regarding Vayne. I want to try to reach the AS cap in game and have a few questions about it.
I am currently a level 22 summoner and wanted to try this before I start playing ranked games. Mainly just for fun, but I am also curious how viable it is and am wondering if anyone has already tried it.
Basically, I ran some numbers and this is what I came up with (I am by no means a math expert, my math is very simplistic, so feel free to let me know what I am missing here). Also I am not looking to be 100% exact with these numbers. Just something close.
Vayne's base AS is 0.658. Now if we just look purely at percentages, you would need roughly an extra 380% to reach the 2.5 attacks per second cap.
Vayne gains 3.1% AS per level which equals an extra 52.7% @ level 18.
Vayne can gain up to 42% AS from runes/masteries.
Itemization: Berserker's + Stinger + 4 Phantom Dancers = 285% AS.
285 + 42 + 52.7 = 379.7% AS at level 18.
Now that I have layed all that out, here are my questions:
1) Is my math at least somewhat correct or close to it? Lol.
2) Assuming my math and itemization is correct, has anyone ever tried this? Was it fun?
3) Is this even somewhat viable? I know someone will probably tell me that this is way to much AS and you're probably right. Throwing in a BT or Edge or Black Cleaver would probably make me pwn a lot harder. But I really just wanna know if I could wreck some face in my level 22, solo queue, non-ranked games with this build.
Any help and correction on this topic would be much appreciated!
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Warrrrax
Senior Member
Your math looks fine.
Im sure people have tried it to some extend. Stacking AS that is.
It isn't really all that viable, mostly because Vayne definitely needs some sustain for early game, and she needs some AD to last hit and harass. Simply put, during laning its very difficult to get in multiple shots and you tend to tumble in, get a shot off and move back. Even during teamfights, etc she tends to shoot and scoot. Tumble in, shoot once or twice, tumble to the side move back, pop ult to stealth, etc. She keeps moving and uses her tumble and knockback. Even when chasing she gets in 1 shot periodically and AS is irrelevant mostly, exceot for faster attk animation.
This type of playstyle lends itself to having lots of AD and crit, and less AS. Silver bolts just isnt enough DPS gain to try to optimize based on AS. Even KOG MAW tends to build AD instead of AS and at level 9 he is adding 120 onhit dmg vs a 2k foe! Thats a crapton of damage, not true dmg, but still.
4 PDs would be 120% crit which is hugely wasteful and the movespeed woulod ahve huuuge diminishing returns.
Instead of this approach maybe try a more moderate one... get your 2 Dorans Blades, vamp scepter, and maybe 1 PD first instead of BFSword/IE. Then the IE followed by 1 more PD, which gets you to 85% crit. Youd then upgrade vamp scepter to executioners blade for maxed crit, high lifesteal without needing bloodthirster.
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Owl Be Back
Senior Member
I've tried it, and it's fun as heck.
it's also SLIGHTLY viable.
I wouldn't do it in ranked, but you can get away with it in normals.
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Simiun
Junior Member
Thanks Warrrrax. You make some really good points. I didn't think much of the play style of Vayne when putting this together. She is really more of a poker than a machine gunner, lol. Which I'm guessing is why most people stack so much AD with her.
Honestly, before coming up with this build, I did have my own build that I was doing (and still am). Granted, I lane with my wife every single game, and she plays a really solid tanky Amumu who gives me all the LHs and kills so I am fed pretty fast (and don't need much life steal because she tanks well and I play safe).
I usually start with the Boots of Swiftness for some early chase downs and then rush PD. The reason I rush PD is because in solo queue at my level I find it hard to get kills once we start having some team fights around mid game. No one seems to understand the concept that the ADC needs kills for the long haul, lol. I find that the extra AS gets me a few free kills simply cause I am attacking faster. After PD I grab a Black Cleaver and then Blood Thirster > Infinity Edge.
Do you think that combo that you recommended, Dorans x2 > Vamp Scepter > PD > IE > PD > Executioners, would be better than what I was doing?
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Simiun
Junior Member
Quote:
Cardboardowl:
I've tried it, and it's fun as heck.
it's also SLIGHTLY viable.
I wouldn't do it in ranked, but you can get away with it in normals.
Lol. Thanks for the reply man. Glad to hear it is fun. That was what I figured. Fun as **** in normals but not for ranked.
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kenoin
Member
Yeah, [ viable, as previously stated not in ranked though ] But lets hope your bottom lane that your facing is stupid or else you're going to have a very hard time trying to make this work. Best rearguards, kenoin.
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Sereg Anfaug
Senior Member
While Vayne does benefit from AS moreso then other ADCs, that doesn't mean you should stack it at the expense of everything else.
Rather, work a little bit more if it into your build then you would with a standard carry. As I don't play Vayne, I can't offer anything more specific then that, but I have played on teams with Vaynes that have gone all out on attack speed, and they were totally unable to play their role. Don't do that, please.
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AdNauseum
Junior Member
Your calculation for attack speed is missing the base multiplier...so you only need 280% aspd from items, masteries and runes total.
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Warrrrax
Senior Member
Quote:
Simiun:
Thanks Warrrrax. Do you think that combo that you recommended, Dorans x2 > Vamp Scepter > PD > IE > PD > Executioners, would be better than what I was doing?
I am not an expert at her, so I can't really vouch for its effectiveness beyond theorycrafting. Did you check Solomid for guides? Im pretty sure they will say the same kind of thing though. Many will rush BFSword to IE in line with her big poke aspect (after a few dorans and possibly a vamp scepter if more sustain is needed or even a wriggles).
I think that is the most effective. But I believe a PD first is doable and fits your concept a lot better. Youd get even more manueverability plus a much better chance at getting your Silver Bolt damage (and applying Ult AD dmg faster). But you really do need some lifesteal and some extra health and some AD for early on laning.
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Eledhan
Senior Member
Quote:
Warrrrax:
I think that is the most effective. But I believe a PD first is doable and fits your concept a lot better. Youd get even more manueverability plus a much better chance at getting your Silver Bolt damage (and applying Ult AD dmg faster). But you really do need some lifesteal and some extra health and some AD for early on laning.
Yeah, pretty much this. Unfortunately, this concept primarily works onKog'Maw, though. Every other ADC (including Vayne) really should be looking for more raw AD or crit damage (meaning Bloodthirster or Infinity Edge) to scale up better. Because Vayne's AS scaling proc requires multiple "procs" to actually trigger, she doesn't get the same level of benefits out of AS like Kog does.
However, rushing a Zeal after 2 Dblades isn't BAD, necessarily, on Vayne, you just are probably better served by getting more AD, since her other two spells scale on that and not on AS. M5's Genja (ADC) is a consistent user of the 2 Dblades & Zeal rush for the mobility and AS it provides, so I guess there are merits to it regardless of champion.
TL/DR - Player preference. If you like the high AS/MS from Zeal/PD, then rush that after you get some dblades. Otherwise, build Vayne similar to the other raw AD scaling ADC's like Graves. | 2,133 | 8,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2015-22 | longest | en | 0.961839 |
https://universalium.academic.ru/98922/critical_point | 1,606,607,931,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141195929.39/warc/CC-MAIN-20201128214643-20201129004643-00690.warc.gz | 532,332,903 | 11,379 | # critical point
critical point
1. Physics. the point at which a substance in one phase, as the liquid, has the same density, pressure, and temperature as in another phase, as the gaseous.
2. Math.
a. (of a function of a single variable) a point at which the derivative of the function is zero.
b. (of a function of several variables) a point at which all partial derivatives of the function are zero.
[1875-80]
* * *
In science, the set of conditions under which a liquid and its vapour become identical.
The conditions are the critical temperature, the critical pressure, and the critical density. If a closed vessel is filled with a pure substance, partly liquid and partly vapour, and the average density equals the critical density, the critical conditions can be achieved. As the temperature is raised, the vapour pressure increases, and the gas phase becomes denser while the liquid expands and becomes less dense. At the critical point, the densities of liquid and vapour become equal, eliminating the boundary between the two.
* * *
▪ phase change
in physics, the set of conditions under which a liquid and its vapour become identical (see phase diagram). For each substance, the conditions defining the critical point are the critical temperature, the critical pressure, and the critical density.
This is best understood by observing a simple experiment. If a closed vessel is filled with a pure substance, partly liquid and partly vapour, so that the average density equals the critical density, the critical conditions can be achieved. As the temperature is raised, the vapour pressure increases, and the gas phase becomes denser. The liquid expands and becomes less dense until, at the critical point, the densities of liquid and vapour become equal, eliminating the boundary between the two phases. If the average density at the start is too low, all the liquid will evaporate before the critical temperature is reached. If the initial average density is too high, the liquid will expand to fill the container.
* * *
Universalium. 2010. | 427 | 2,061 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2020-50 | latest | en | 0.927652 |
http://mrburkemath.blogspot.com/2019/08/august-2019-algebra-1-regents-part-1.html | 1,579,266,945,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250589560.16/warc/CC-MAIN-20200117123339-20200117151339-00144.warc.gz | 114,078,413 | 29,180 | Friday, August 23, 2019
August 2019 Algebra 1 Regents Part 1
The following are some of the multiple questions from the recent August 2019 New York State Common Core Algebra I Regents exam.
Omitted images will be added soon.
August 2019 Algebra I, Part I
Each correct answer is worth up to 2 credits. No partial credit. Work need not be shown.
1. Bryan’s hockey team is purchasing jerseys. The company charges \$250 for a onetime set-up fee and \$23 for each printed jersey. Which expression represents the total cost of x number of jerseys for the team?
The start-up fee is paid once, so it doesn't get a variable. It's the cost even if 0 jerseys are made, so it's the y-intercept of a graph. The \$23 is a repeated cost per jersey, so it gets a variable.
2. Which table represents a function?
Choices (1), (2) and (3) each repeat an x value with a different y value. That is not allowed in a function.
3. Which expression is equivalent to 2(x2 - 1) + 3x(x - 4)?
Answer: (4) 5x2 - 12x - 2
2(x2 - 1) + 3x(x - 4)
2x2 - 2 + 3x2 - 12x
5x2 - 12x - 2
You might've used process of elimination part-way through this problem.
4. The value of x that satisfies the equation 4/3 = (x + 10) / 15 is
The quick way: 3 goes into 15 five times, and 4 * 5 = 20 = x + 10. Then x = 10.
or
4/3 = (x + 10) / 15
(3)(x + 10) = (4)(15)
3x + 30 = 60
3x = 30
x = 10
5. Josh graphed the function f(x) = -3(x - 1)2 + 2. He then graphed the function g(x) = -3(x - 1)2 - 5 on the same coordinate plane. The vertex of g(x) is
Answer: (1) 7 units below the vertex of f(x)
The vertex of f(x) is (1, 2). The vertex of g(x) is (1, -5).
Therefore the vertex of g(x) is 7 units lower than f(x).
6. A survey was given to 12th-grade students of West High School to determine the location for the senior class trip. The results are shown in the table below.
Niagara Falls Darien Lake New York City Boys 56 74 103 Girls 71 92 88
To the nearest percent, what percent of the boys chose Niagara Falls?
Sum the number of boys by adding the three responses: 56 + 74 + 103 = 233.
Divide the number for Niagara Falls by 233, and make it a percent: 56 / 233 * 100% = 24%
Note that you don't need to use any of the data for the girls.
7. Which type of function is shown in the graph below?
Basically, a definition question. Linear functions and absolute value functions are straight lines, not curves.
A square root bends the other way, with smaller increases in y and x gets larger.
8. The expression 16x2 - 81 is equivalent to
Answer: (3) (4x - 9)(4x + 9)
The Difference of Squares Rule: when you have the difference between two terms which are each perfect squares, take the square root of each term and write the conjugates. (That is, write the two terms twice, once with a plus between them, and once with a minus.)
9. The owner of a landscaping business wants to know how much time, on average, his workers spend mowing one lawn. Which is the most appropriate rate with which to calculate an answer to his question?
Time per lawn: time can be measured in hours and lawns in the unit measurement.
Lawns per day doesn't give the amount of time for one lawn.
10. A ball is thrown into the air from the top of a building. The height, h(t), of the ball above the ground t seconds after it is thrown can be modeled by h(t) = -16t2 + 64t + 80. How many seconds after being thrown will the ball hit the ground?
Put the equation in your graphing calculator, and check the table. It has zeroes at -1 and 5. Reject -1 because negative time makes no sense.
The long way:
-16t2 + 64t + 80 = 0 -- Divide by -16 to get smaller numbers.
t2 - 4t - 5 = 0
(t - 5)(t + 1) = 0
t = 5 or t = -1, reject -1
11. Which equation is equivalent to y = x2 + 24x - 18?
Answer: (1) y = (x + 12)2 - 162
You don't have to complete the square because choices are given.
Actually, you wouldn't have to complete the square even without the choices.
Put the graph into the calculator and look at the table. Where is the vertex? It is at (-12, -162).
You can also see that (-12, 126), (12, 162) and (12, 126) are not even points on the parabola.
Next method: the vertex is on the Axis of Symmetry, so x = -b / (2a) = -24/(2*1) = -12.
Eliminate choices (3) and (4).
Then y = (-12)2 + 24(-12) - 18 = -162.
12. When (x)(x - 5)(2x + 3) is expressed as a polynomial in standard form, which statement about the resulting polynomial is true?
(x)(x)(2x) = 2x3, which is be the leading term, with the highest exponent. This also means that the degree is 3, and not 2.
The constant will be (-5)(3) = -15, not 2. And there will be more than these 2 terms.
13. The population of a city can be modeled by P(t) = 3810(1.0005)7t , where P(t) is the population after t years. Which function is approximately equivalent to P(t)?
(1.0005)7t = (1.00057)t.
1.0005 to the 7th power is approximately 1.0035.
The rest of the equation remains unchanged.
14. The functions f(x) and g(x) are graphed on the set of axes below.
For which value of x if f(x) =/= g(x)?
The graphs intersect at x = -2, x = -1 and x = 2. At x = 3, g(3) = 0, but f(3) is off the top of the graph.
15. What is the range of the box plot shown below?
The range is 8 - 1 = 7.
The difference 6 - 4 (which is 2) would be the Interquartile Range, which is different.
16. Which expression is not equivalent to 2x2 + 10x + 12?
Answer: (3) (2x + 3)(x + 4)
Not to be a broken record, but you could put each of these into the graphing calculator and see if they give you the same graph. However, in this case, that would actually take more time.
Just looking at the choices, you should immediately see that (2) and (4) are equivalent to each other because (2x + 6) is the same as 2(x + 3). So neither of those in the answer.
Likewise, there is no way that both (1) and (3) could be equivalent: switching the 3 and 4 will result in a different middle term. So check those:
(2x + 4)(x + 3) = 2x2 + 6x + 4x + 12 = 2x2 + 10x + 12. Check!
(2x + 3)(x + 4) = 2x2 + 8x + 3x + 12 = 2x2 + 11x + 12. Wrong!
17. The quadratic functions r(x) and q(x) are given below
x r(x) -4 -12 -3 -15 -2 -16 -1 -15 0 -12 1 7
q(x)=x2 + 2x - 8
The function with the smallest minimum value is
Answer: (3) r(x), and the value is -16
The minimum for r(x) is -16.
Graph q(x) and find the vertex. Or use the Axis of Symmetry, x = -2/(2(1)) = -1.
Evaluate q(-1) = (-1)2 + 2(-1) - 8 = 1 - 2 - 8 = -9.
The function with the smallest minimum value is r(x), and that value is -16 (which occurs when x = -2, but that is NOT what is being asked).
18. A child is playing outside. The graph below shows the child's distance, d(t), in yards from home over a period of time, t, in seconds.
Which interval represents the child constantly moving closer to home?
Answer: (1) 0 < t < 2
If the child is moving toward home at a constant rate, the distance is decreasing steadily.
That happens between 0 and 2 seconds. at you're accidentally get the "right" answer to appear by mistake.
19. If a1 = 6 and an = 3 + 2(an - 1)2, the a2 equals
a2 = 3 + 2(a2 - 1)2 = 3 + 2(a1)2
= 3 + 2(6)2 = 3 + 2(36) = 3 + 72 = 75
20. The length of a rectangular patio is 7 feet more than its width, w. The area of a patio, A(w), can be represented by the function
Answer: (2) A(w) = w2 + 7w
Length = w + 7. Area = length * width = (w + 7)(w) = w2 + 7w.
21. A dolphin jumps out of the water and then back into the water. His jump could be graphed on a set of axes where x represents time and y represents distance above or below sea level. The domain for this graph is best represented using a set of
Forget about the above and below the water. They asked for the domain, which is the x value, which in this case represents time. Negative time makes no sense.
I don't like this problem because it leaves out 0, which is not positive, but should be part of the domain. However, the answer key tells me that it is not part of the domain.
22. Which system of linear equations has the same solution as the one shown below?
x - 4y = 10
x + y = 5
Answer: (1) 5x = 10, x + y = 5
Change x + y = 5 into y = 5 - x
That makes the first equation x - 4(5 - x) = -10
and then x - 20 + 4x = -10
So 5x - 20 = -10
And 5x = 10, which is choice (1).
23. Which interval represents the range of the function h(x) = 2x2 - 2x - 4?
It's a parabola that opens upwards, so the upper boundary is infinity. The lower boundary is the minimum point at the vertex. Just looking at the choices, the answer should be obvious once you realize what the incorrect answer is.
You can graph this and look for the vertex.
Or use the Axis of Symmetry: x = -b/(2a) = -(-2)/(2(2)) = 0.5, which is NOT the range.
h(.5) = 2(.5)2 - 2(.5) - 4 = -4.5, which is part of the range, so use [ not (.
24. What is a common ratio of the geometric sequence whose first term is 5 and third term is 245?
The second term must be n/5 = 245/n, so n2 = (5)(245)
n = SQRT(1225) = 35
35 / 5 = 7, and 245 / 35 = 7
The Common ratio is 7.
End of Part I
How did you do? | 2,803 | 8,961 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-05 | latest | en | 0.883128 |
https://www.askiitians.com/forums/Mechanics/q35-the-acceleration-of-two-objects-are-5m-s2-and_185833.htm | 1,611,253,190,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703527224.75/warc/CC-MAIN-20210121163356-20210121193356-00112.warc.gz | 655,605,244 | 34,940 | ×
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```
Q35. The acceleration of two objects are 5m/s2 and 20m/s2 ,if masses of both the objects would be combined and a force of 50N is applied to it ,the what will be the acceleration
Q35. The acceleration of two objects are 5m/s2 and 20m/s2 ,if masses of both the objects would be combined and a force of 50N is applied to it ,the what will be the acceleration
```
3 years ago
John Sharon Sandeep
41 Points
``` I dont think the question is correct as neither the force exerted on the objects are given nor their mass.But incase 50N is the force applied individually, following is the answer🔜Since F=m/a , m1=F/a1 =50N/5ms-1 =>10kg m2=F/a2=50N/20ms-1 =>2.5kgTherefore total mass on combining=m1+m2=10+2.5=12.5kg Thus,required acceleration=a=F/m => a=50N/12.5kg=4ms-2😆
```
3 years ago
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## Other Related Questions on Mechanics
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### Course Features
• 101 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution
### Course Features
• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions | 467 | 1,675 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2021-04 | latest | en | 0.797112 |
https://www.instructables.com/id/Getting-around-in-Minecraft/ | 1,566,476,165,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027317113.27/warc/CC-MAIN-20190822110215-20190822132215-00227.warc.gz | 862,119,166 | 25,157 | # Getting Around in Minecraft
33,636
63
18
There are a few ways to find your way around in the game. You can just wander around, keep track of your coordinates, craft up a boat, or even zoom around in a Minecart. Check out this collection of ways to find your way around in Minecraft.
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Coordinates
Coordinates are very handy for finding your way around assuming you remember the coordinates of where you were so you can find your way back later.
To see your coordinates, you need to pull up the Debug screen which is F3 on the computer version. The Debug screen gives you a lot of information, but we are going to concentrate on the coordinate related points. The points are based on an origin point.
What you are going to use from the Debug Screen:
• X - Coordinate : shows how far East (positive number) or West (negative number) you are
• Z - Coordinate : shows how far South (positive number) or North (negative number) you are
• Y - Coordinate : shows how above or below Sea Level (Level 63) you are (this does not go negative)
Direction - You can tell what direction you are facing by looking at the f: number
• 0 = South
• 1 = West
• 2 = North
• 3 = East
Biome - the b: tells you which Biome you are in; the screenshot shows I am in "Plains"
Using Coordinates:
• I find it most handy to record the X and Z of certain locations you want to keep in mind (home, lava pool, jungle, Mooshroom Biome, etc)
• If you are able to walk exactly straight in one direction, only the X or Z coordinates will change, this is very difficult and most likely you they will both change as you go
## Step 2: Maps
Maps are kind of troublesome to use. They are mostly handy for keeping track of one area of the game.
When you crate a map, you will be able to see yourself as a white arrow shape wandering around that map. If you leave the area of the map, you can create another map and now on that map you will be able to see your white arrow shape. If you go back into the area of the first map, it will no longer work. Only your current map will show your exact location as you move around.
In order to use a map, you have to complete it. To complete it, hold your empty map like you would any resource and click to use it. You can watch as the map creates itself. Once it is active, the empty map icon will look the same except it will have black squiggles on it.
Each map you complete will have a number. Even if you delete a previous numbed map, the maps numbers will continue to go up.
To cover more area on a single map, you can zoom it out by combining it with more paper. You can zoom out a map 4 levels besides your start level meaning there are 5 levels the map can show. At the lowest level of the map you everything will show up in a 1 to 1 scale. So for each box on the map, it will be a box in the game.
Crafting:
• 3 Sugar Canes = 3 Pieces of Paper
• 4 Iron Ingots + 1 Redstone = 1 Compass
• 8 Pieces of Paper + 1 Compass = 1 Empty Map
• 8 Pieces of Paper + 1 Completed Map = 1 Zoomed Out Map
• Completed Map + Blank Map = Cloned Map
• ## Step 3: Boats
Boats are amazing for covering great distances on water since they go faster than you walk or swim. The downside to them is they are extremely fragile. You can destroy a boat by hitting a lily pad, hitting a squid or just going too close to shore while going to fast. If you are boating near show, it is wise to slow down.
A boat destroyed by crashing it will drop 2 Sticks and 3 Wood Planks, unless you destroy a boat by hitting it, then it will drop a boat.
When not riding your boat, you are going to want to destroy it to get a boat or "dock it". The best way to dock it is to get close to land, preferably with a side already blocked in. Then completely block it in on all sides. It should stay there. Do not dock a boat by digging into the shore, driving the boat into the hole you have created and then blocking it in. If you do, the water will flow into the hole until you block it off, then it will stop flowing and you will get the boat suck on land. You can get it back on water again, but it is easier to dock it in still water.
Crafting:
• 5 Wood Planks = 1 Boat
## Step 4: How to Use Rails and Minecarts
Rains and Minecarts are very handy for use while mining in Minecraft, but they can also be used as transportation. To make them useful you are going to need to learn more about powered tracks. I talk about them in my Instructable How to Mine in Minecraft, but I will cover them more here as well.
When using rails, you can use normal rails by themselves but you have to direct them like when you walk and they won't go very fast unless you are going downhill
Common Trails of Redstone Rails
• Power with a Redstone Block, RedstoneTorch, Redstone in combination with a Lever or Button or Restone Torch, or just a Lever or Button; using a lever will turn the rail on and off, a button will turn it on temporarily and a Redstone Torch will leave it on
• Rails connected together will power one another if they are the same; i.e. a powered Powered Rail will activate any other Powered Rails it touches but will not power an Activator or Detector Rail (this is not true with a Detector Rail since its purpose is to power/activate other rails)
• Powered Rails and Activator Rails will show up red when placed next to a Redstone power source; a Detector Rail will show up red once activated by a Minecart and so will the rail it activates
Powered Rails
• Most helpful of the special rails
• Are used to speed up and stop Minecarts : Speed up when Activated; Stop when Deactivated
• Can be powered from above, below or from any side
• Combine with Detector Rails for 1 Way (one Detector rail on one end of the Powered Rails) or 2 Way (one Detector rail on each end of the Powered Rails) travel
• Optimal spacing is 1 every 38 blocks - but this only works if the minecart has gathered speed; when the minecart first starts going they need to be closer otherwise it won't reach the next powered rail; to get it started try not to space them more than 8 or so regular rails apart (to figure out the spacing the best thing to do is just activate the Powered Rail with the Minecart on it and just see how far it can go, then place a Powered Rail as far as you can that it will still reach
Detector Rails
• Is a pressure plate and can be used to activate Redstone or Powered Rails
• Is finicky on slopes
Activator Rails
• Is used with special Minecarts such as those containing TNT or a Hopper
• When it is off it acts like a normal rail unlike a Powered Rail
• When combined with a TNT Minecart it will light the TNT
• When combined with a Hopper Minecart, it turns off the Hopper
• If something is in the cart, such as you or a mob, it will be ejected when going over an Activator Rail
Crafting Minecarts
• 5 Iron Ingots in a "u" = Minecart
• Minecart + Chest = Minecart with Chest
• Minecart + Furnace = Minecart with Furnace
• Minecart + Hopper = Minecart with Hopper
• Minecart + TNT = Minecart with TNT
Crafting Rails:
• 6 Iron Ingots + 1 Stick = 16 Rails
• 6 Gold Ingots + 1 Stick + 1 Redstone = 6 Powered Rails
• 6 Iron Ingots + 1 Stone Pressure Plate + 1 Redstone = 6 Detector Rails
• 6 Iron Ingots + 1 Redstone Torch + 2 Sticks = 6 Activator Rails
Other Supplies
• 1 Stick + 1 Block of Cobblestone = 1 Lever
• 1 Stone | 1 Wooden Plank = 1 Button
## Recommendations
• ### Lamps Class
9,590 Enrolled
## 18 Discussions
O yea u can use soulsand for dock. Boats will NOT break on it.
O and I have an artillery piece too, so I don't need golems.
This is a nice post. Actually, there are users who find it hard to screenshot Minecraft which i think is so easy. This was my problem before until I found this very helpful article http://screenshot.net/minecraft-screenshot.html
Does the empty map has to be 8 paper and a compass! Does the empty map fill in its self?? | 1,969 | 8,038 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2019-35 | longest | en | 0.934583 |
https://www.jiskha.com/display.cgi?id=1231786006 | 1,501,210,402,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549436321.71/warc/CC-MAIN-20170728022449-20170728042449-00135.warc.gz | 814,315,944 | 4,132 | # math
posted by .
a restraurant has 11 booths and 12 tables, write the ratio of booths to tables as a fraction
• math -
11/23? is that what you mean?
• math -
yes that the answer I got but was no sure thanks
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Post a New Question | 531 | 2,123 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2017-30 | longest | en | 0.95745 |
https://omargamaleldeen.com/2018/08/08/653-two-sum-iv-input-is-a-bst/ | 1,679,651,876,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945279.63/warc/CC-MAIN-20230324082226-20230324112226-00466.warc.gz | 502,963,032 | 21,162 | # 653. Two Sum IV – Input is a BST
653. Two Sum IV – Input is a BST
```/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean findTarget(TreeNode root, int k) {
ArrayList data = new ArrayList();
traverers(root, data);
for (int i = 0 ; i < data.size() ; i++)
for (int j = 0 ; j < data.size() && i!=j ; j++)
if (data.get(i) + data.get(j) == k)
return true;
return false;
}
private void traverers(TreeNode root, ArrayList data) {
if (root != null) {
data.add(root.val);
traverers(root.left, data);
traverers(root.right, data);
}
}
}
``` | 201 | 683 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2023-14 | latest | en | 0.173791 |
https://socratic.org/questions/how-do-you-solve-4m-3-2 | 1,638,424,043,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00594.warc.gz | 600,312,974 | 5,904 | # How do you solve 4m-3=2?
Dec 28, 2016
$\textcolor{red}{\text{m = 1.25}}$
#### Explanation:
$\textcolor{b r o w n}{\text{Add 3 to both sides:}}$
$4 m - \cancel{3} = 2$
$\textcolor{w h i t e}{a a a} + \cancel{3}$ $\textcolor{w h i t e}{a} + 3$
3 cancels out on the left side, and when you add 3 to 2 on the right you get 5:
$4 m = 5$
$\textcolor{b l u e}{\text{Divide both sides by 4 to get m by itself.}}$
(cancel"4"m)/cancel"4"=(5)/4
Now you are just left with $m$ on the left side and 1.25 on the right side.
Thus,
$\textcolor{p u r p \le}{\text{m = 1.25}}$ | 235 | 571 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 10, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2021-49 | latest | en | 0.683301 |
https://thvinhtuy.edu.vn/basic-geometry-concepts-video-lessons-diagrams-examples-step-by-step-solutions-a2bbikdc/ | 1,718,790,821,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861817.10/warc/CC-MAIN-20240619091803-20240619121803-00347.warc.gz | 508,383,186 | 13,240 | # Basic Geometry Concepts (video lessons, diagrams, examples, step-by-step solutions)
Geometric Plane Shapes For Kids – Primary Vocabulary
Geometric Plane Shapes For Kids – Primary Vocabulary
Related Pages
2-D and 3-D Shapes
More Geometry Lessons
These lessons introduces basic geometry terms including: points, lines, line segments, midpoints, rays, planes and space. Building upon these basic ideas, we look into some introductory geometry concepts.
Basic Geometry Geometric Terms Fundamental Concepts of Geometry Angles Geometric Theorems Geometry Worksheets Math Worksheets
The following table gives some geometry concepts, words and notations. Scroll down the page for examples, explanations and solutions.
We may think of a point as a “dot” on a piece of paper or the pinpoint on a board. In geometry, we usually identify this point with a number or letter. A point has no length, width, or height – it just specifies an exact location. It is zero-dimensional.
Every point needs a name. To name a point, we can use a single capital letter. The following is a diagram of points A, B, and M:
We can use a line to connect two points on a sheet of paper. A line is one-dimensional. That is, a line has length, but no width or height. In geometry, a line is perfectly straight and extends forever in both directions. A line is uniquely determined by two points.
Lines need names just like points do, so that we can refer to them easily. To name a line, pick any two points on the line.
The line passing through the points A and B is denoted by
A set of points that lie on the same line are said to be collinear. Pairs of lines can form intersecting lines, parallel lines, perpendicular lines and skew lines.
Because the length of any line is infinite, we sometimes use parts of a line. A
line segment connects two endpoints. A line
segment with two endpoints A and B is denoted by
.
A line segment can also be drawn as part of a line.
The midpoint of a segment divides the segment into two segments
of equal length. The diagram below shows the midpoint M of the line segment . Since M is the midpoint, we know that the lengths AM = MB.
A ray is part of a line that extends without end in one direction. It starts from one endpoint and extends forever in one direction.
A ray starting from point A and passing through B is denoted by
Planes are two-dimensional. A plane has length and width, but no height, and extends infinitely on all sides. Planes are thought of as flat surfaces, like a tabletop. A plane is made up of an infinite amount of lines. Two-dimensional figures are called plane figures.
All the points and lines that lie on the same plane are said to be coplanar.
A plane.
Space is the set of all points in the three dimensions – length, width and height. It is made up of an infinite number of planes. Figures in space are called solids.
Figures in space.
This video explains and demonstrates the fundamental concepts (undefined terms) of geometry: points, lines, ray, collinear, planes, and coplanar. The basic ideas in geometry and how we represent them with symbols.
A point is an exact location in space. They are shown as dots on a plane in 2 dimensions or a dot in space in 3 dimensions. It is labeled with capital letters. It does not take up any space.
A line is a geometric figure that consists of an infinite number of points lined up straight that extend in both directions for ever (indicated by the arrows at the end). A line is identified by a lower case letter or by two points that the line passes through. There is exactly 1 line through two points. All points on the same line are called collinear. Points not on the same line are noncollinear.
Two lines (on the same plane) are either parallel or they will meet at a point of intersection.
A line segment is a part of a line with two endpoints. A line segment
starts and stops at two endpoints.
A ray is part of a line with one endpoint and extends in one direction forever.
A plane is a flat 2-dimensional surface. A plane can be identified by 3 points in the plane or by a capital letter. There is exactly 1 plane through three points. The intersection of two planes is a line.
Coplanar points are points in one plane.
An angle consists of two rays with a common endpoint. The two rays are called the sides of the angle and the common endpoint is the vertex of the angle.
Each angle has a measure generated by the rotation about the vertex. The measure is determined by the rotation of the terminal side about the initial side. A counterclockwise rotation generates a positive angle measure. A clockwise rotation generates a negative angle measure. The units used to measure an angle are either in degrees or radians.
Angles can be classified base upon the measure: acute angle, right angle, obtuse angle, and straight angle.
If the sum of measures of two positive angles is 90°, the angles are called complementary.
If the sum of measures of two positive angles is 180°, the angles are called supplementary.
Examples:
The Opposite Angle Theorem (OAT)
When two straight lines cross, opposite angles are equal.
The Angle Sum of a Triangle Theorem
The interior angles of any triangle have a sum of 180°.
The Exterior Angle Theorem (EAT)
Any exterior angle of a triangle is equal to the sum of the opposite interior angles.
Parallel Lines Theorem (PLT)
Whenever a pair of parallel lines is cut by a transversal
a) corresponding angles are equal (PLT-F)
b) alternate angles are equal (PLT-Z)
c) interior angles have a sum of 180° (PLT-C)
Try the free Mathway calculator and
problem solver below to practice various math topics. Try the given examples, or type in your own | 1,238 | 5,699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-26 | latest | en | 0.907286 |
https://www.sawaal.com/partnership-questions-and-answers/a-began-a-business-with-rs-85-000-he-was-joined-afterwards-by-b-with-ks-42-500-for-how-much-period-d_2129 | 1,713,527,184,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296817398.21/warc/CC-MAIN-20240419110125-20240419140125-00698.warc.gz | 895,022,026 | 14,114 | 91
Q:
# A began a business with Rs. 85,000. He was joined afterwards by B with Ks. 42,500. For how much period does B join, if the profits at the end of the year are divided in the ratio of 3 : 1 ?
A) 4 months B) 5 months C) 6 months D) 8 months
Explanation:
Suppose B joined for x months . Then, ( 85000 * 12 )/(42500 * x) = 3. or x = (85000 * 12) / (42500 * 3) = 8.
So, B joined for 8 months.
Subject: Partnership
Q:
Two partners M and N buy a car. M pays his share of 3/7th of the total cost of the car. M pays Rs. 31,540 less than N. What is the cost of the car?
A) Rs. 2,32,680 B) Rs. 2,03,175 C) Rs. 2,20,780 D) Rs. 1,85,780
Explanation:
Filed Under: Partnership
Exam Prep: Bank Exams
6 1348
Q:
Tapan, Ravi and Trisha shared a cake. Tapan had 1/4 of it, Trisha had 2/3 of it and Ravi had the rest. What was Ravi's share of the cake?
A) 4/7 B) 1/12 C) 1/6 D) 2/6
Explanation:
Filed Under: Partnership
Exam Prep: Bank Exams
8 1311
Q:
Mohan invested Rs. 100000 in a garment business. After few months, Sohan joined him with Rs. 40000. At the end of the year, the total profit was divided between them in ratio 3 : 1. After how many months did Sohan join the business?
A) 3 B) 2 C) 4 D) 5
Explanation:
Filed Under: Partnership
Exam Prep: Bank Exams
11 1502
Q:
The central Government granted a certain sum towards flood relief to 3 states A, B and C in the ratio 2 : 3 : 4. If C gets Rs. 400 Crores more than A, what is the share of B?
A) Rs. 400 crore B) Rs. 200 crore C) Rs. 600 crore D) Rs. 300 crore
Explanation:
Filed Under: Partnership
Exam Prep: Bank Exams
4 1158
Q:
Sunil and Gopal are partners in a business. Sunil invests Rs.20,000 for 9 months and Gopal invests Rs. 30,000 for 12 months. Find the share of Sunil from the total profit of Rs. 60,000.
A) Rs.15000 B) Rs.18000 C) Rs.25000 D) Rs.20000
Explanation:
Filed Under: Partnership
Exam Prep: Bank Exams
2 1290
Q:
P starts a business with Rs 14000. After 8 months Q joins P with Rs 8000. After 3 years, what will be the ratio of the profit of P and Q?
A) 12 : 5 B) 9 : 4 C) 14 : 9 D) 15 : 8
Explanation:
Filed Under: Partnership
Exam Prep: Bank Exams
6 1758
Q:
C and D started a business by investing Rs 4000 and Rs 5000 respectively. After 3 months, C withdraws Rs 1000 while D invest Rs 1000 more after 6 months of the starting of business. At the end of the year C’s share is Rs 3900. What will be the total profit (in Rs) at the end of year?
A) 7800 B) 10500 C) 8400 D) 12200
Explanation:
Filed Under: Partnership
Exam Prep: Bank Exams , CAT
2 1572
Q:
X starts a business with Rs 80000. After 6 months Y joins X with Rs 100000. After 2 years, what will be the ratio of profit of X and Y?
A) 16 : 15 B) 4 : 5 C) 8 : 9 D) 14 : 15 | 942 | 2,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2024-18 | latest | en | 0.941456 |
https://whatisconvert.com/253-oil-barrels-in-tablespoons | 1,660,115,697,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571150.88/warc/CC-MAIN-20220810070501-20220810100501-00359.warc.gz | 564,882,203 | 7,400 | # What is 253 Oil Barrels in Tablespoons?
## Convert 253 Oil Barrels to Tablespoons
To calculate 253 Oil Barrels to the corresponding value in Tablespoons, multiply the quantity in Oil Barrels by 10751.999999964 (conversion factor). In this case we should multiply 253 Oil Barrels by 10751.999999964 to get the equivalent result in Tablespoons:
253 Oil Barrels x 10751.999999964 = 2720255.9999908 Tablespoons
253 Oil Barrels is equivalent to 2720255.9999908 Tablespoons.
## How to convert from Oil Barrels to Tablespoons
The conversion factor from Oil Barrels to Tablespoons is 10751.999999964. To find out how many Oil Barrels in Tablespoons, multiply by the conversion factor or use the Volume converter above. Two hundred fifty-three Oil Barrels is equivalent to two million seven hundred twenty thousand two hundred fifty-six Tablespoons.
## Definition of Oil Barrel
A barrel is one of several units of volume included fluid barrels (UK beer barrel, US beer barrel), dry barrels, oil barrel, etc. Since medieval times, the size of barrel has been used with different meanings around Europe, from about 100 liters to above 1000 in special cases. Modern barrels are made of aluminum, stainless steel, and different types of plastic, such as HDPE. Now in most countries, the barrels are replaced by SI units. However, prices per barrel in USD are commonly used.
## Definition of Tablespoon
In the United States a tablespoon (abbreviation tbsp) is approximately 14.8 ml (0.50 US fl oz). A tablespoon is a large spoon used for serving or eating. In many English-speaking regions, the term now refers to a large spoon used for serving, however, in some regions, including parts of Canada, it is the largest type of spoon used for eating. By extension, the term is used as a measure of volume in cooking.
## Using the Oil Barrels to Tablespoons converter you can get answers to questions like the following:
• How many Tablespoons are in 253 Oil Barrels?
• 253 Oil Barrels is equal to how many Tablespoons?
• How to convert 253 Oil Barrels to Tablespoons?
• How many is 253 Oil Barrels in Tablespoons?
• What is 253 Oil Barrels in Tablespoons?
• How much is 253 Oil Barrels in Tablespoons?
• How many tbsp are in 253 bbl?
• 253 bbl is equal to how many tbsp?
• How to convert 253 bbl to tbsp?
• How many is 253 bbl in tbsp?
• What is 253 bbl in tbsp?
• How much is 253 bbl in tbsp? | 597 | 2,390 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-33 | latest | en | 0.921721 |
https://web2.0calc.com/questions/stuck-on-these-questions | 1,558,714,966,000,000,000 | text/html | crawl-data/CC-MAIN-2019-22/segments/1558232257660.45/warc/CC-MAIN-20190524144504-20190524170504-00047.warc.gz | 670,574,799 | 5,797 | +0
# Stuck on these questions
0
68
1
+62
1.Let $f(x) = \frac{2}{\sqrt{x}}$. What is the domain of $f$? Express your answer with interval notation.\
2.Let $f(x) = \frac{3x - 7}{x + 1}.$Find the domain of $f$. Give your answer using interval notation.
Mar 1, 2019 | 99 | 266 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2019-22 | longest | en | 0.6829 |
http://www.physicsforums.com/showthread.php?t=99080 | 1,368,875,622,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368696382360/warc/CC-MAIN-20130516092622-00024-ip-10-60-113-184.ec2.internal.warc.gz | 655,847,264 | 6,847 | ## Irreducible polynomial on polynomial ring
How would I prove that $x^2+1$ is irreducible in $Z_p[x]$, where p is an odd prime of the form 3+4m.
I know that for it to be rreducible, it has to have roots in the ring. So $x^2=-1 (mod p)$. Or $x^2+1=k(3+4m)$, for some k. I tried induction on m, but it does not work because [itex}x^2+1[/itex] is only reducible on $Z_p[x]$ if p is prime, which is not the case for all m. Apperently, there exists a two-line solution.
Any tips would be appreciated.
PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
Hint: Fermat's little theorem and this lemma: if R is a commutative ring with identity, and a in R is invertible, then a^n=1 and a^m=1 => a^gcd(n,m)=1. | 257 | 874 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2013-20 | latest | en | 0.90718 |
http://notesforfree.com/2017/09/15/practical_xii_chromatography/ | 1,553,611,203,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205534.99/warc/CC-MAIN-20190326135436-20190326161436-00281.warc.gz | 152,750,923 | 26,665 | # Chromatography
AIM: To separate the coloured components present in a mixture of red and green ink by ascending paper chromatography and find their Rf values.
THEORY:
In this type of chromatography, a special adsorbent paper (Whatman filter paper) is used. Moisture adsorbed bon this Whatman filter paper acts as the stationary phase and the solvent acts as the mobile phaseThe mixture to be separated is spotted at one end of the paper. This paper is then developed in a particular solvent by placing the paper in a gas jar, taking care that the spot is above the solvent. The solvent rises due to capillary action and the components get separated out as they rise up with the solvent at different rates. The developed paper is called a chromatogram.
Rf (retention factor) values are then calculated, which is the ratio of the distance moved by the component to the distance moved by the solvent front.
Rf = Distance traveled by the component
Distance traveled by the solvent front
OBSERVATIONS AND CALCULATIONS: (ON THE BLANK PAGE, USING A PENCIL)
Component Distance Travelled by Different Components Distance Travelled by Solvent Rf Value Red dr= 2cm ds= 4cm 0.5 Green dg= 3.8cm ds = 4cm 0.95
RESULT: (ON RULED SIDE
The components of the mixture (red and green colour) separate in the form of spots lying between the origin line and solvent front.
Rf(green) = dg/ds = 3.8/4 =0.95
Rf(red) = dr/ds = 2/4= 0.5
Precautions
• The strip should not touch the walls of the jar.
• Do not disturb the jar after putting the strip in it.
• Allow the colours to seperate before pulling the strip out.
You can also get Class XII Practicals on BiologyPhysics, and Physical Education.
## 4 Replies to “Chromatography”
Yo dawg this shit is real cool kuku
2. kriti dogra says:
thanks
3. Harsh says:
Awesome
Really helpful and saved my marks.
4. Sana Jahangir says:
It was really very helpful for me!
Please post the experiments of Salt Analysis of class XI. | 495 | 1,966 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2019-13 | latest | en | 0.911681 |
https://www.primepuzzles.net/puzzles/puzz_667.htm | 1,620,554,500,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988966.82/warc/CC-MAIN-20210509092814-20210509122814-00301.warc.gz | 983,639,000 | 3,951 | Problems & Puzzles: Puzzles
Puzzle 667 Puzzle 77 This puzzle has received one solution in 1999 a better one in 2009 and no more after that. Perhaps it's time to get one more solution now. After all it's almost christmas! Q. Improve the solutions to Puzzle 77
Contributions came from Seiji Tomita and Jan van Delden
***
Seiji wrote:
Q3:
Solutions for R=7, only bottom rows.
Ending digit 1: 3631 2731 991 2251 11311 8011 11551, Total=149278
Ending digit 3: 823 463 2503 1663 9463 2383 8263, Total=101644
Ending digit 7: 2687 1907 47 227 2207 14867 3407, Total=89756
Ending digit 9: 1709 4289 509 929 269 449 24029, Total=78782
Solutions for R=8.
Ending digit 1: 26641 8941 3361 541 6481 12781 41761 51421, Total=489546
Ending digit 3: 22123 1723 1123 163 283 11083 42643 22963, Total=334188
Ending digit 7: 6277 8737 9397 13537 7717 17377 10837 26017, Total=428652
Ending digit 9: 499 15619 3019 5659 2179 23059 3739 19819, Total=312354
***
Jan wrote:
Q3:
Solutions with R=8, bottom rows (not necessarily the least assignation):
Ending digit 1: 14051 22751 4691 10271 17891 6911 11411 12671
Ending digit 3: 34673 36473 35393 19913 19793 35993 14753 4073
Ending digit 7: 11617 14197 4177 6277 12097 6037 4657 13477
Ending digit 9: 13789 25609 1069 7129 3229 8329 4909 10009
***
Records | Conjectures | Problems | Puzzles | 538 | 1,353 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2021-21 | longest | en | 0.39839 |
http://www.etedal.net/2014/03/dh-parameters.html | 1,580,128,923,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251700675.78/warc/CC-MAIN-20200127112805-20200127142805-00307.warc.gz | 210,209,761 | 15,128 | ## Sunday, March 23, 2014
### There are a few rules to consider in choosing the coordinate system:
1. the $z$-axis is in the direction of the joint axis
2. the $x$-axis is parallel to the common normal: $x_n = z_n \times z_{n - 1}$
If there is no unique common normal (parallel $z$ axes), then $d$ (below) is a free parameter.
3. the $y$-axis follows from the $x$- and $z$-axis by choosing it to be a right-handed coordinate system.
Once the coordinate frames are determined, inter-link transformations are uniquely described by the following four parameters:
• $\theta\,$: angle about previous $z$, from old $x$ to new $x$
• $d\,$: offset along previous $z$ to the common normal
• $r\,$: length of the common normal (aka $a$, but if using this notation, do not confuse with $\alpha$). Assuming a revolute joint, this is the radius about previous $z$.
• $\alpha\,$: angle about common normal, from old $z$ axis to new $z$ axis
## "Standard" DH Parameters
Following the DH standard you must provide 4 numbers that define the orientation of the ith link with respect to the i-1th link. "Standard" DH convention assumes that the ith coordinate frame is at the i+1 joint. (joint 1 axes 0, joint 2 axes 1 ...)
2. (Link parameter)(a)The second number represents the length (in meters) along xi of the common perpendicular between zi-1 and zi.
3. (Joint parameter)(theta)The third number represents the angle (in radians) between xi-1 and xi about zi-1.
4. (Joint parameter)(d)The fourth number represents the distance (in meters) along axis zi-1 between the origin of the i-1th coordinate frame and the point where the common perpendicular intersects axis zi
## "Modified" DH Parameters (also called Craig's convention)
Following the modified DH standard, you must provide 4 numbers that define the orientation of the ith link with respect to the i-1th link. Unlike the "standard" DH convention, the "modified" DH convention assumes that the ith coordinate frame is at the i joint. (joint 1 axes 1,joint 2 axes 2 ...) | 518 | 2,017 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 21, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.484375 | 3 | CC-MAIN-2020-05 | latest | en | 0.851236 |
https://math.answers.com/math-and-arithmetic/What_is_n_negative_3_equals_negative_8 | 1,720,966,436,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514580.77/warc/CC-MAIN-20240714124600-20240714154600-00078.warc.gz | 350,677,758 | 47,934 | 0
# What is n negative 3 equals negative 8?
Updated: 6/9/2024
Wiki User
11y ago
n - 3 = -8
n - 3 + 3 = -8 + 3
lenpollock
Lvl 16
1mo ago
Wiki User
11y ago
n - 3 = 8 (add three to both sides)
n = 11
Earn +20 pts
Q: What is n negative 3 equals negative 8?
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N = -4 and 3/8
### What is n minus negative 3 equals negative 4?
If: n--3 = -4 Then: n = -7
2=n
### How do you multiply 3 negative integers?
=when you multiply negative integers the answer is going to be negative because n-multiply by n- equals n+ but when yu multiply n+ by n- it equals n- so you basicly multiply the numbers and the answer would be ah negative.==when you multiply negative integers the answer is going to be negative because n-multiply by n- equals n+ but when yu multiply n+ by n- it equals n- so you basicly multiply the numbers and the answer would be ah negative.=
3n=24 n=8
### What is N if n plus 3 over 8 equals negative 4?
n = -35(n+3)/8 = -4Multiply both sides by 8.8(n+3)/8 = -4 x 8Cancel 8 on the left-hand side.8(n+3)/8 = -4 x 8Simplify.n+3 = -32Subtract 3 from both sides.n = -32 - 3Simplify.n = -35
n=24
n = 3/8
### -17x equals -204 21 equals -7n -143 equals -11x -16 plus x equals -15 n-8 equals -10?
For the equation -17x = -204, divide both sides by -17 to get x = 12. For the equation 21 = -7n, divide both sides by -7 to get n = -3. For the equation -143 = -11x, divide both sides by -11 to get x = 13. For the equation -16 + x = -15, add 16 to both sides to get x = 1. For the equation n - 8 = -10, add 8 to both sides to get n = -2.
24
### What is 16-n equals 8?
16-n equals 8 = 8
n=-4/3 | 567 | 1,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2024-30 | latest | en | 0.867117 |
http://www.ck12.org/book/CK-12-Geometry-Concepts/r8/section/12.2/ | 1,493,467,649,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917123491.68/warc/CC-MAIN-20170423031203-00369-ip-10-145-167-34.ec2.internal.warc.gz | 486,109,969 | 42,342 | # 12.2: Rotation Symmetry
Difficulty Level: At Grade Created by: CK-12
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Practice Rotation Symmetry
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What if you were asked to consider the presence of symmetry in nature? The starfish, below, is one example of symmetry in nature. Draw in the center of symmetry and the angle of rotation for this starfish. After completing this Concept, you'll be able to answer questions like these.
### Guidance
Rotational Symmetry is when a figure can be rotated (less that 360\begin{align*}360^\circ\end{align*}) and it looks the same way it did before the rotation. The center of rotation is the point at which the figure is rotated around such that the rotational symmetry holds. Typically, the center of rotation is the center of the figure. Along with rotational symmetry and a center of rotation, figures will have an angle of rotation, that tells us how many degrees we can rotate a figure so that it still looks the same. In general, if a shape can be rotated n times, the angle of rotation is 360n\begin{align*}\frac{360^\circ}{n}\end{align*}. Then, multiply the angle of rotation by 1, 2, 3..., and n\begin{align*}n\end{align*} to find the additional angles of rotation.
#### Example A
Determine if the figure below has rotational symmetry. Find the angle and how many times it can be rotated.
The pentagon can be rotated 4 times and show rotational symmetry. Because there are 5 lines of rotational symmetry, the angle would be 3605=72\begin{align*}\frac{360^\circ}{5}= 72^\circ\end{align*}. Note that the 5th rotation would be 360\begin{align*}360^\circ\end{align*} and so does not count for demonstrating rotational symmetry.
#### Example B
Determine if the figure below has rotational symmetry. Find the angle and how many times it can be rotated.
The N\begin{align*}N\end{align*} can be rotated once. The angle of rotation is 180\begin{align*}180^\circ\end{align*}.
#### Example C
Determine if the figure below has rotational symmetry. Find the angle and how many times it can be rotated.
The checkerboard can be rotated 3 times. There are 4 lines of rotational symmetry, so the angle of rotation is 3604=90\begin{align*}\frac{360^\circ}{4}=90^\circ\end{align*}. It can also be rotated 180\begin{align*}180^\circ\end{align*} and 270\begin{align*}270^\circ\end{align*} and it will still look the same.
Watch this video for help with the Examples above.
#### Concept Problem Revisited
The starfish has rotational symmetry of 72\begin{align*}72^\circ\end{align*}. Therefore, the starfish can be rotated 72,144,216\begin{align*}72^\circ, 144^\circ, 216^\circ\end{align*}, and 288\begin{align*}288^\circ\end{align*} and it will still look the same. The center of rotation is the center of the starfish.
### Vocabulary
Rotational symmetry is present when a figure can be rotated (less than 360\begin{align*}360^\circ\end{align*}) such that it looks like it did before the rotation. The center of rotation is the point a figure is rotated around such that the rotational symmetry holds. The angle of rotation that tells us how many degrees we can rotate a figure so that it still looks the same. In general, if a shape can be rotated n times, the angle of rotation is 360n\begin{align*}\frac{360^\circ}{n}\end{align*}.
### Guided Practice
Find the angle of rotation and the number of times each figure can rotate.
1.
2.
3.
1. The parallelogram can be rotated twice. The angle of rotation is 180\begin{align*}180^\circ\end{align*}.
2. The hexagon can be rotated six times. The angle of rotation is 60\begin{align*}60^\circ\end{align*}.
3. This figure can be rotated four times. The angle of rotation is 90\begin{align*}90^\circ\end{align*}.
### Practice
1. If a figure has 3 lines of rotational symmetry, it can be rotated _______ times.
2. If a figure can be rotated 6 times, it has _______ lines of rotational symmetry.
3. If a figure can be rotated n\begin{align*}n\end{align*} times, it has _______ lines of rotational symmetry.
4. To find the angle of rotation, divide 360\begin{align*}360^\circ\end{align*} by the total number of _____________.
5. Every square has an angle of rotation of _________.
Determine whether each statement is true or false.
1. Every parallelogram has rotational symmetry.
2. Every figure that has line symmetry also has rotational symmetry.
Determine whether the words below have rotation symmetry.
1. OHIO
2. MOW
3. WOW
4. KICK
5. pod
Find the angle of rotation and the number of times each figure can rotate.
Determine if the figures below have rotation symmetry. Identify the angle of rotation.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Center of Rotation In a rotation, the center of rotation is the point that does not move. The rest of the plane rotates around this fixed point.
Rotation A rotation is a transformation that turns a figure on the coordinate plane a certain number of degrees about a given point without changing the shape or size of the figure.
Rotation Symmetry A figure has rotational symmetry if it can be rotated less than $360^\circ$ around its center point and look exactly the same as it did before the rotation.
Symmetry A figure has symmetry if it can be transformed and still look the same.
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Subjects: | 1,405 | 5,593 | {"found_math": true, "script_math_tex": 20, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 1, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2017-17 | latest | en | 0.870532 |
http://list.seqfan.eu/pipermail/seqfan/2015-December/015851.html | 1,643,118,587,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304835.96/warc/CC-MAIN-20220125130117-20220125160117-00695.warc.gz | 39,905,046 | 3,005 | # [seqfan] Re: Organization of dispersion-arrays.
Antti Karttunen antti.karttunen at gmail.com
Tue Dec 15 17:34:32 CET 2015
On Tue, Dec 15, 2015 at 6:14 PM, Frank Adams-Watters <franktaw at netscape.net>
wrote:
> If I understand this correctly, what you are here calling the
> "complementary" sequence is what I have been calling the "ordinal
> transform" of the sequence. I often find it easier to think of the related
> sequences as transforms of each other rather than by looking at the
> associative array.
>
[A bit tired now, so I might make a mistake here, but ...]: I have assumed
that when you write about two sequences that are mutually ordinal
transforms of each other, then it is question of the row and column indices
of such arrays. So it is actually the point
(B) "The column index and the row index" in my mail, what we are talking
about here, not the point (C) quoted below?
So for example, if I understand your terminology right, sequences
https://oeis.org/A254111
and https://oeis.org/A254112 are ordinal transforms of each other?
Best regards,
Antti
>
> Whatever sequence you feed to the ordinal transform - even a
> non-mathematical sequence such as a sequence of words - the result will be
> a sequence of positive integers with the property that for any k and n > 0,
> in the first n terms of the sequence, k will occur at least as many times
> as k+1. This property is characteristic of this transform, since if you
> apply the transform twice to a sequence with this property, you get the
> original sequence back. This is easy to see in terms of the array: you are
> just transposing it.
>
> If the original sequence is fractal - that is, invariant under an upper
> trim: removing the first occurrence of each value in the sequence gives you
> back the same sequence - then its ordinal transform will be invariant under
> the lower trim: remove the 1's from the sequence, and subtract 1 from every
> remaining term; and vice versa, of course.
>
> I guess my main point is that, when complementary sequences are in the
> database, I would like to see a note that they are ordinal transforms of
> each other, whether in a comment, cross-ref, or formula.
>
Yes, certainly.
>
>
>
> -----Original Message-----
> From: Antti Karttunen <antti.karttunen at gmail.com>
> To: Sequence Fanatics <seqfan at list.seqfan.eu>
> Sent: Tue, Dec 15, 2015 2:22 am
> Subject: [seqfan] Organization of dispersion-arrays.
>
> C) First few rows and first few columns. Depending on the orientation of
> the array (which of two transposes it is), the topmost row or the leftmost
> column is the complementary sequence to the sequence whose dispersion this
> array is, and correspondingly, the edgemost sequence on other axis gives
> the iterates of the dispersion sequence itself.
>
> | 704 | 2,794 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.125 | 3 | CC-MAIN-2022-05 | latest | en | 0.932946 |
rayosle-telephone.com | 1,642,720,184,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320302706.62/warc/CC-MAIN-20220120220649-20220121010649-00276.warc.gz | 545,484,883 | 19,545 | Home
# Credit card number format
### Immobilien, Wohnungen, Häuser und Gewerbe ImmoScout2
1. Baufinanzierung für Jedermann - Auch ohne Eigenkapital zum Eigenhei
2. 25€ Startguthaben - Visa, Mastercard, Prepaid usw. Kostenlos im Vergleich. VISA oder Mastercard, Top Kreditkarten - Jetzt vergleichen & abschließen
3. Payment card numbers are composed of 8 to 19 digits, as follows: a six or eight-digit Issuer Identification Number (IIN), the first digit of which is the major industry identifier (MII) a variable length (up to 12 digits) individual account identifier. a single check digit calculated using the Luhn algorithm
4. A credit card number is the string of numeric digits that identifies the credit card. It's usually (but not always) displayed on the front of the card, and it generally (but not always) features 15 or 16 digits
5. ed mathematically based on the Luhn algorithm
To successfully display a 16-digit credit card number in full, you must format the number as text. For security purposes, you can obscure all except the last few digits of a credit card number by using a formula that includes the CONCATENATE, RIGHT, and REPT functions. Display credit card numbers in ful Most credit card number can be validated using the Luhn algorithm, which is more or a less a glorified Modulo 10 formula! The Luhn Formula: Drop the last digit from the number. The last digit is what we want to check against; Reverse the numbers; Multiply the digits in odd positions (1, 3, 5, etc.) by 2 and subtract 9 to all any result higher than 9; Add all the numbers togethe Card Number Card Type Issuing Country Expiry Date CVV2; 4111 1111 4555 1142 security code optional: Classic: NL: 03/2030: 737: 4988 4388 4388 4305: Classic: ES: 03/2030: 737: 4166 6766 6766 6746: Classic: NL: 03/2030: 737: 4646 4646 4646 4644: Classic: PL: 03/2030: 737: 4000 6200 0000 0007: Commercial Credit: US: 03/2030: 737: 4000 0600 0000 0006: Commercial Debit: US: 03/2030: 737: 4293 1891 0000 000 A valid credit card number is developed by the formulation of ISO/IEC 7812 which contains two different parts. They are the numbering system and application and registration procedures Credit Card Numbers. I have a form. On the form is a text box for credit card numbers. I set the character limit to 16, and I set the format to number with no decimal points and no commas. So that when the numbers are entered into the field it will look like a credit card number. My question is: Is there a way to automatically space out numbers. So that when someone inputs all 16 numbers every.
### Kreditkarte mit 25€ Startguth
The proper format of a valid credit card number starts with the major industry identifier or MII, which differentiates each credit card number in accordance to their industries. A Valid credit card number should be: A 16 digit number The Luhn algorithm or Luhn formula, also known as the modulus 10 or mod 10 algorithm, named after its creator, IBM scientist Hans Peter Luhn, is a simple checksum formula used to validate a variety of identification numbers, such as credit card numbers, IMEI numbers, National Provider Identifier numbers in the United States, Canadian Social Insurance Numbers, Israeli ID Numbers, South.
### Payment card number - Wikipedi
My solution in the end, was to use one field. Then every 4 digits, it adds a space. Therefore it creates less potential confusion, yet also allows you to clearly view the credit card numbers, in the same format as the card. The potential issues that arose are: Different number of credit card numbers, e.g. 16 or 1 Format credit card number. How to format and validate a credit card number with spaces between each 4 digit while typing: \$ ('.creditno').keyup (function () { cc = \$ (this).val ().split (-).join (); cc = cc.match (new RegExp ('. {1,4}\$|. {1,4}', 'g')).join (-); \$ (this).val (cc); })
### Anatomy of a Credit Card: Cardholder Name, Number, Network
• One particularly tricky piece of credit card information is the card number, which is typically a non-sensical 16-digit long string where even a single typo will result in a card validation error
• A number in the format of a credit card number. A number that passes the checksum. A keyword or an expiration date in the right format. You can use these confidence levels (or match accuracy) in your rules
• Card numbers can be typed by users: As one single long number; As four sets of four numbers separated by dashes; As four sets of four numbers separated by spaces; You have two choices in how to deal with this. Inform users of what format to type their card number in and restrict their input
To successfully display a 16-digit credit card number in full, you must format the number as text. For security purposes, you can obscure all except the last few digits of a credit card number by using a formula that includes the CONCATENATE, RIGHT, and REPT functions. Display credit card numbers in full . Select the cell or range of cells that you want to format. How to select a cell or a. Credit Card Account Numbers Format. Your credit card account numbers are known as your Primary Account Numbers. Credit Card Account Numbers are referred to as being in the PAN format for short. All credit card account numbers as well as debit cards are in this PAN format. Merchant cards have their own format. American Express cards have 15 digits while other credit and debit cards have 16 digits Because custom number formats are designed to work primarily with numbers, you cannot create a custom number format that stores more than 15 digits. For example, you cannot use the following format to store a 16-character credit card ID as a number: ####-####-####-#### If you type the number 1111222233334444 in a cell that uses the. John Batdorf is correct here - several card carriers have more or less than the standard 16 digits. For example, American Express has 15 digits, arranged in a format of 0000 000000 00000. Other international cards such as Maestro can have as few as 12 and as many as 19 digits. You may need to take the card carrier into account when performing the display. See Credit Card Type: Credit Card Number: American Express: 378282246310005: American Express: 371449635398431: American Express Corporate: 378734493671000: Australian BankCard: 5610591081018250: Diners Club: 30569309025904: Diners Club: 38520000023237: Discover: 6011111111111117: Discover: 6011000990139424: JCB: 3530111333300000: JCB: 3566002020360505: MasterCard: 5555555555554444: MasterCar
The most important deviation is the 15-digit AMEX number which uses a 4-6-5 chunking. Any 19-digit credit card number generally uses 4-4-4-4-3 chunking From the rightmost digit of your card number, double every other digit. If the doubled digit is larger than 9 (ex. 8 * 2 = 16), subtract 9 from the product (16-9 = 7) A credit card number is the long set of digits displayed across the front or back of your plastic credit card. It is typically 16 digits in length, often appearing in sets of four. Sometimes it can be as long as 19 digits, and it is used to identify both the credit card issuer and the account holder. Credit card numbers are not randomly assigned
The front of your credit card has a lot of numbers -- here's an example of what they might mean. Illustration by Rosaleah Rautert Although phone companies, gas companies and department stores have their own numbering systems, ANSI Standard X4.13-1983 is the system used by most national credit-card systems American Express credit card account numbers are 15 digits in lengths, and generally start with either 34 or 37. An input-validation regex for 15-digit American Express card numbers, e.g. 371449635398431 4569403961014710 5191914942157165 370341378581367 38520000023237 6011000000000000 3566002020360505 1234566660000222 4569403961014710 5191914942157165 370341378581367 38520000023237 6011000000000000 3566002020360505 1234566660000222 Credit card number is the card unique identifier found on payment cards
Validating credit card numbers is the ideal job for regular expressions. They're just a sequence of 13 to 16 digits, with a few specific digits at the start that identify the card issuer It allows you to validate all payment testing scenarios such as credit card number length, format, type, issuing network etc. Features. Generate thousands of fake / dummy credit card numbers & details using our free bulk generator tool. Download card details in three formats: JSON, XML & CSV. Each card is generated with completely random information including name, address, zipcode and country. BobbyScon's answer points you to the industry standard PCI rules but, the anatomy of a credit card number is as follows. First Digit is major ID number (3 = AMEX/Diners Club, 4 = Visa, 5 = Mastercard, 6 = Discover, etc Validating a credit card number BEFORE it gets sent off to the processor is not the same as storing it. A lot of people do it to avoid being charged by the processor for sending an invalid number. How many people here can say that they can not only safely store and transmit this information as a 3rd party and pay for HTTPS and also the Certificates that will be needed. Believe me, anyone who. Credit card number generator features. Luhn algorithm checked; With credit card number validator; 100% Valid Credit Card Numbers; Major industry identifier added; Generate up to 999 values per click! Used for data testing and verification purposes; 100% FREE to generate Credit Cards; With random 3-digit security code; With random names, addres
Here are some format of some well-known credit cards. American Express :- Starting with 34 or 37, length 15 digits. Visa :- Starting with 4, length 13 or 16 digits. MasterCard :- Starting with 51 through 55, length 16 digits. Discover :- Starting with 6011, length 16 digits or starting with 5, length 15 digits A credit card number consist of complex formulation of ISO/IEC 7812 which has 2 different parts - the numbering system and application and registration procedures. It consist of prefix digit Major Industry Identifier, 6-digit issuer identification number or IIN, 7-digit personal account number Credit Card Type IIN ranges Credit Card Number; Discover: 6011, 622126-622925, 644-649, 65: 6011594786717005: MasterCard: 2221-2720: 5229657264125429: American Express: 34, 37: 37431347670181 React Credit Card Number Formatter. Adapter component that helps in building a payment card Form. Formats credit card numbers to readable format: #### #### #### #### Validates card numbers with luhn check; Blocks invalid input for MM, YY and CVV fields; Agnostic about styling, unopinionated about the UI, works with all UI Librarie
Best online tool to check and verify Credit Card and Debit Cards by Bank Identification Number (BIN/IIN). BIN Credit Card Checker is very useful in commercial business for fraud prevention, especially in online store. BIN search information lookup have over 1 000 000 unique Card and Debit Card numbers in the database Digits 7 to final number minus 1 indicate the individual account identifier (the last is the checksum). Issuer identification number (IIN) The first six digits of a card number identify the institution that issued the card to the card holder
### What Do the Numbers on Your Credit Card Mean
• Make sure that the Credit Card or Debit Card Number follows the proper format. The length is in between 13 to 19 characters and contains only numbers and no space in between. Example : 371449635398431 / 4532421174341278 / 5569755825672968 Validation Required. About Credit Card / Debit Card Number Checker Tool. Credit Card / Debit Card Number Checker tool is designed to check the validity of.
• Pattern: Credit card number Credit card number is one of the patterns that you can select on the Match panel. Use this pattern to match a series of digits that looks like a credit card number from one or more of the world's major credit card issuers. This example shows how you can use the credit card pattern to look for credit card numbers. You can find this example as Pattern.
• Credit Card Generator's primary role is data verification and software testing. It is a complete legal tool and binds by all laws laid down by the government. It is a flexible and legal online CC Generator for generating credit card numbers. If a website is fake or illegal, use this card generator to save yourself from financial fraud. It also allows you to authenticate all payment testing details such as credit card number length, format, type, issuing network, etc
• As a credit card number can be of maximum 19 digits, and as the first 6 digits are issuer identifier and the last digit is the check digit, hence, an account number in a credit card can be of 12..
• For checking a credit card is valid or not, the following are the validations we have to be sure for declaring the result. A credit card's number must have 13 to 16 digits, it must start with the following digits. All the visa cards start from 4 All the master cards start from
• On an actual credit card, the digits of the embossed card number are usually placed into groups of four. That makes the card number easier for humans to read. Naturally, many people will try to enter the card number in the same way, including the spaces, on order forms
• This is a sample Visa credit card number with CVV and expiration date from Idaho central c.u bank. 4511625822501241 07/22 042 credit visa classic idaho central c.u. US Buhl ID Judy Mansfield 4034 Blue Cascade Court. 4452522001128360 04/22 434 credit visa classic andrews f.c.u. US York PA Patricia Fonzi 812 McKenzie St . This is a fake cc also known as a dummy credit card. From Mastercard.
### Display numbers as credit card numbers - Exce
• Credit Card Generator includes MII. The Maestro credit card generator is entirely free to generate credit card numbers. You can download card in JPEG format. You can download bulk Maestro credit card details in CSV, XML and JSON
• Fortunately, credit card numbers are created in a way that allows for some basic verification. This verification does not tell you if funds are available on the account, and it certainly doesn't tell whether or not the person submitting the order is committing credit card fraud. In fact, it's possible that the card number is mistyped in such a way that it just happens to pass verification.
• The IIN makes up the first six digits of all credit or debit cards issued by American Express, followed by the primary account number (PAN) and a check digit. For more information, contact American Express at (800) 444-2450 or by visiting www.americanexpress.com
When we say a dummy credit card number we are basically implying that these credit card number are created with the same numbering formulation as of those a real credit card number which can be easily done by simply assigning particular credit card number prefixes. Such as the following: 4 for Visa - Banking and financia Card Number Format Type of Card Country Issued ; 377852 37785 2XXXX XXXXX: CREDIT: Australia 516310 5163 10XX XXXX XXXX: CREDIT: Australia 516319 5163 19XX XXXX XXXX: CREDIT: Australia 516361 5163 61XX XXXX XXXX: DEBIT: Australia 516390 5163 90XX XXXX XXXX: DEBIT: Australia 377852 37785 2XXXX XXXXX: CREDIT: Australia 516310 5163 10XX XXXX XXXX: CREDIT: Australia 516319 5163 19XX XXXX XXXX. First, let us see some examples of valid and invalid credit card numbers with our conditions applied to it for a python program to validate a given credit card number. 92136258797157867 #17 digits in card number -> Invalid; 7123456789123456 ->Valid; 3123-7754-9978-2343 ->Valid; 4997-5624-9832-2211 Starting with digit 4 -> Invalid ; 9675 - 7756-8864-9075 contains space in between ->Invalid.
### Credit Card Number Generator & Validator - FreeFormatter
1. Our credit card generator tools work in a similar form, like how credit card issuers make their credit cards. You can check available credit/debit card Bank Identification Number (BIN) details using our BIN Checker tool to identify the card-issuing details that issued the card to the cardholder
2. Credit card type can be deduced from the card number. The first digits of a card number identifies the organisation that issued the card. For the purposes of online payment, you currently mostly.
3. CardJS is a very simple, clean, credit card form for your website. Includes number formatting, validation and automatic card type detection. 2. Skeuocard . Github. Skeuocard is a JavaScript plugin that allows you to progressively enhance a credit card form providing a skeuomorphic interface. When you begin entering your card number, Skeuocard attempts to match it to an accepted card type. Once.
4. Place all the credit card logos inside the number input by default, then as the user types in the first two numbers, all the card logos disappears except for the one that corresponds to the input. 6. EXPIRY DATE. Most credit cards display their expiry dates in the format MM/YY (month and year). Some may include the full year, in an YYYY format. When designing the expiry date input, we wanted.
5. Each American Express card number begins with 34 and 37 like that of Visa card and Mastercard format with 5 and 4 numbers respectively. Meanwhile, with number 5, MasterCard begins as contrast, number 4 is started with by Visa Card and number 6 is started with by Discover's
6. Credit card numbers are usually 16 digits. When you enter a number greater than 15 digits, Excel rounds it off and you get zeros padding the end of your number. Enter the number . 987654321098765432. in Excel and you get this. 987654321098765000. If you have a large number like this that's causing a problem in Excel, there's usually one of two reasons: You are an astrophysicist and you.
7. In this tutorial, we are going to create a simple Credit Card Form using HTML and CSS3.We'll work with Google Fonts to use the custom font (Roboto) for this form. You are free to integrate this form into your website
Each line in the input field must have the following format: Credit card name, Bank identification number prefix, Credit card length Use the comma ( , ) to separate the Credit card name, Bank identification number prefix and Credit card length. Use the pipe ( | ) to separate individual values (45|46|47) Use the dash ( - ) to specify a range of values (45-47). Example Custom credit card number. jquery-creditcard-formatter. jQuery library that transforms credit card numbers via aesthetic formatting based on defined credit cards. Examples. User inputs credit card number: 371449635398431 (American Express) Result: 3714 4963539 8431. User inputs credit card number: 4111111111111111 (Visa) Result: 4111 1111 1111 111
What's In A Number At a glance, a credit card number just appears to be a sequence of digits. You may have noticed that the major processing providers have their own prefixes. Visa cards all start with a 4, MasterCard with 5, Discover with 6, and American Express are 3 (and 15 digits instead of 16). Further, financial institutions will have. Credit card numbers follow certain patterns. A credit card number must have between 13 and 16 digits. It must start with: 4 for Visa cards; 5 for Master cards; 37 for American Express cards; 6 for Discover cards. The problem can be solved by using Luhn algorithm. Luhn check or the Mod 10 check, which can be described as follows (for.
Credit card numbers and expiration dates are stored in the OPERA database tables in encrypted (AES256) format. All user log entries containing a credit card number are also encrypted. Masking. As applied to credit card information, masking involves displaying Xs for all numbers of a credit card, except for the last four digits of the credit card number. The expiration date is also masked with. If s1 + s2 ends in zero then the original number is in the form of a valid credit card number as verified by the Luhn test. For example, if the trial number is 49927398716: Reverse the digits: 61789372994 Sum the odd digits: 6 + 7 + 9 + 7 + 9 + 4 = 42 = s1 The even digits: 1, 8, 3, 2, 9 Two times each even digit: 2, 16, 6, 4, 18 Sum the digits of each multiplication: 2, 7, 6, 4, 9 Sum the last. There's a hidden mathematical pattern in your credit card (also works for debit cards or ATM cards). In this video I will teach you a mathematical magic tric.. Credit cards numbers generated follows the rule of Luhn algorithm and ISO/IEC 7810 numbering standard, credit cards generated comes with all the necessary fake security details such as fake name, CVV, Country origin, and much more! Using creditcardinfo.com is the easiest way to get a credit card number for your needs. To start generating Valid Credit cards simply follow instructions below.
### Credit Card Generator Credit Card Numbers Generator
• Structure of a credit card number Credit cards throughout the world follow the same standard when it comes to numbering. It's the International Standards Organization (ISO/IEC 7812-1:1993) and the American National Standards Institute (ANSI X4.13) that has drawn up the specifications for credit card numbering system
• For example, a credit card number beginning with 7 is probably an oil company or gas station card. Step 2 Count the number of digits in the credit card. Cards with 14 digits are typically Diner's Club and cards with 15 digits are typically American Express or older JCB credit cards. Most other major types of credit cards have 16 digits. Step 3 Match the first few digits of the credit card.
• Der Card Validation Code (CVC) (auch Card Verification Value (CVV), Card Verification Number (CVN), Card Security Code (CSC), Card Code Verification (CCV), Kartenprüfnummer (KPN)) oder Sicherheitscode ist ein Sicherheitsmerkmal auf Kreditkarten.. Die Prüfnummer soll die Nutzung von gefälschten oder gestohlenen Kreditkartenangaben erschweren, da sich hiermit feststellen lassen soll, ob eine.
• Generate Credit Card Numbers with Complete Details . Easily generate a valid credit card numbers in just few clicks. You can now generate your own valid credit card numbers with CVV, country origin, issuing network (such as Visa, Master Card, Discover, American Express and JCB), account limit, and expiry date.. GetCreditCardInfo.com aims to deliver a valid credit card numbers to everyone.
• Credit Card formatter in Javascript. 19 November 2013 27 comments JavaScript. Peterbe.com Credit Card formatter in Javascript. Home Archive About Contact. Menu Home Archive About Contact Search. Mind that age! This blog post is 8 years old! Most likely, its content is outdated. Especially if it's technical. I looked around for Javascript libs that do automatic input formatting for credit card.
Credit Card numbers without any hyphens and within a Word or PDF document are not detected, when sent attached to an e-mail. Note: The very same documents I tested successful in the portal! Here an example of the numbers: 4532 1753 6071 1112 4539 5385 7425 5825 5325 3256 9519 6624 4532 4220 6922 9909 4532 0065 1968 5602 4556 0072 1294 7415 5527 1247 5046 7780 5299 1561 5689 1938. When you. Credit Card Vault. The CC Vault Function is within the property interface configuration and can be set at the property level. To eliminate the storage of credit card numbers in OPERA, Unique IDs (encrypted credit card keys) will be used to replace any credit card numbers; thereafter, these unique IDs will be used for any of the guest's transactions at the property No need to change your form. Card works as a drop in addition to your current credit card form. No need edit input names or HTML — you can leave everything as is. All pure CSS, HTML, and Javascript. There are no images used. The credit cards and logos are all hand-coded with HTML and CSS. 100% free and open sourc The following are credit card numbers in a valid format: American Express: 3400 0000 0000 009: Carte Blanche: 3000 0000 0000 04: Discover: 6011 0000 0000 0004: Diners Club: 3852 0000 0232 37: enRoute: 2014 0000 0000 009: JCB: 3530 111333300000: MasterCard: 5500 0000 0000 0004: Solo: 6334 0000 0000 0004: Switch: 4903 0100 0000 0009: Visa: 4111 1111 1111 1111 : Laser: 6304 1000 0000 0008: Select. For example, let's say I want to encrypt my credit card number. In principle this is no problem: card numbers are just numbers, after all. Assuming that we want a general technique*, I could simply represent my 16-digit VISA card number as an integer between 0 and 9,999,999,999,999,999. In binary this value will require up to 54 bits to store (since lg 10,000,000,000,000,000 = 53.1508495.
### Credit Card Numbers (PDF Forms) - Acrobat Answer
• This template demonstrates form with credit card payment. It includes card number, expiration date, cv code and card owner names
• Never store electronic track data or the card security number in any form. While you may have a business reason for storing credit card information, processing regulations specifically forbid the storage of a card's security code or any track data contained in the magnetic strip on the back of a credit card. The card security number, called by many acronyms including CVV2, CID, and CSC.
• Credit card forms are also part of the process when someone wants to sign up in getting a certain card. Thus, our list of templates of credit card forms come in handy. These are ideal for those employees who are working on banks, credit firms, and so on. Each contains different content to fit every individual's preference
• Apple in their Credit Card Payment Form detects the type of credit card you're using and makes it a little bit easier as you don't need to choose your type from a traditional list. The responsiveness on the form is also important as it helps people focus attention. When you start to type your credit card number the right icon will remain colorful, while the other will fade into gray scale
### Credit Card Generator - CC Generato
1. In this content, we're going to talk about free Visa Credit Card Generator tool & the process to use it. Visa Credit Card Generator. Nowadays, it has become effortless and convenient to generate Visa Credit Card numbers entirely using fake details which include fake name, house address, phone number, expiry date and security details like the 3- digit security code which is also known as CVV
2. On credit cards, the checksum takes the form of a check digit. In a typical 16-digit credit card number, the first six digits identify the institution that issued the card. The next nine digits identify the individual account associated with the card. The last digit, the 16th, is the check digit. Credit card issuers plug the first 15 digits into a mathematical formula called the Luhn.
3. Credit Card Formatter is a very small jQuery plugin for a credit card input that formats numbers and tells you the detected credit card type. 3. Card. Card will take any credit card form and make it the best part of the checkout process. Everything is created with pure CSS, HTML, and Javascript. 4. Skeuocard . Skeuocard is a re-think of the way we handle credit card input on the web. It.
4. Fast User-friendly Form Validation Engine - Supreme Validation. HTML5 Form Validator For Bootstrap 5 - jbvalidator. Feature-rich Input Validation & Mask Plugin - InnerFormValidation. Minimal Form Validator With jQuery And Regex - jQuery validateMini. Credit Card Form With Cool Interactions. Easy Form Validation Plugin For Bulma Framewor
5. Genuine card information cannot be used in test mode. Instead, use any of the following test card numbers, a valid expiration date in the future, and any random CVC number, to create a successful payment. Each basic test card's billing country is set to U.S. If you need to create test card payments using cards for other billing countries, use.
### Luhn algorithm - Wikipedi
So for our test credit card number 378282246310005, we apply a modulus of 10 to the result from step two, like this: 60 MOD 10 = 0. The modulus operation returns 0, indicating that the credit card. Your invoice payment has been declined and you are receiving the message 'Invalid Credit Card Number' on your Dashboard. Resolution. When you select Pay Now in the Billing section of your Dashboard, you may receive the message Invalid Credit Card Number. This indicates a problem with the information entered for the card. Double-check the credit card number, expiration date, and CVV. If you.
Credit Card Payment Form Company Name . Full Corporate Name Please. Help Us Avoid Confusion! Enter Dollar Amount . USD Credit Card First Name : Last Name : Credit Card Number : Security Code : Expiration Month : Expiration Year : Billing Address Street Address : Street Address Line 2 : City : State / Province : Postal / Zip Code : Country : Upload Signed Documents Here (If you Want, Not. Credit Card Payment Form Company Name * Email * example@example.com . Invoice Number(s) * Separate with comma, e.g. 1234, 2123, 3126 . Payment Amount . CAD Total of all invoices listed above. Chips make it harder for thieves to use stolen credit card numbers. If your card has a chip, use it whenever possible by inserting your card instead of swiping. The chip adds a single-use code to every transaction, which makes stolen data less useful. Preventing fraud can keep costs down for everybody, and it means you're less likely to have to replace cards and update card numbers after.
### forms - Splitting credit card number fields into four
1. Click Credit card types and set up information for each type of credit card that you accept. For more information, see Credit card types (form). Close the Credit card types form. In the Payment services form, select the Default processor for new credit cards check box, and then close the form. Test the credit card authorization process
2. Test Credit Card Numbers; Test Credit Card Numbers. Below is a list of test credit card numbers to be used with the test environment. They will pass Luhn's MOD-10 algorithm, but have no actual accounts associated with them. Any expiration date from the future will work. American Express. 3782 82246 310005 3714 49635 398431 Corporate: 3787 34493 671000 Diners Club. 3852 0000 023237 3056 9309.
3. APR assumes that you draw your entire credit limit in one transaction using your card for a period of one year, and repay in 12 equal instalments, the principal plus the respective interest amounts (if your credit limit has not yet been finalised, we have assumed that it will be €1,500), and Government stamp duty on the card is charged, the total amount you would have to pay would be €.
4. A credit card generator creates fake but valid credit card numbers using the same algorithms used by banks and card networks to issue their cards. Discard Card Generator can also use the BIN code (bank identifier) of any bank in the world. Since generated cards are not actually connected to any bank account, they have no funds and can't be used to purchase goods or services, but they are.
5. If the cardholder includes a credit card number on any form or document other than the Credit Card Payment Form or submits this form electronically via EFS-Web, the United States Patent and Trademark Office will not be liable in the event that the credit card number becomes public knowledge. Title : Microsoft Word - PTO-2038_revised 12-04-13 Author: turanga Created Date: 12/5/2013 11:06:31 AM.
6. Credit Card Number - Also known as Primary Account Number (PAN). The one thing to point out is that credit card numbers have some unique properties. For instance, the first number of the card number will vary depending on the type of card. Card: Starting Number: Visa: 4: MasterCard: 5: Discover: 6: A great article about credit card numbers is Anatomy of Credit Card Numbers. Another notable.
7. This example shows how to validate a credit card number to protect against typos. The program first verifies that the number has 16 digits. It then checks the digits starting from the end. If a digit's position in the number is odd (numbered starting with digit 1 at the end), the program adds its value to a checksum value. If the digit's position is even, the program doubles it and adds the.
### javascript - Format credit card number - Stack Overflo
1. Description: Makes the element require a credit card number. Return true if the value is a valid credit card number. Works with text inputs. Part of the additional-methods.js file. Note: The algorithm used can't verify the validity of the number - it is just an integrity check. As with any other clientside validation, you have to implement the same or better validation on the serverside.
2. Lines are open 7am to 11pm, seven days a week. Further details on your chargeback rights and additional protection for credit card purchases under Section 75 of the Consumer Credit Act are available in the Travel FAQs. If it's been less than 8 weeks, please continue to complete the form below. What happens after making a clai
3. Credit cards: are issued by MBNA Limited. Registered Office: Cawley House, Chester Business Park, Chester CH4 9FB. Registered in England and Wales under company number 02783251. Authorised and regulated by the Financial Conduct Authority. MBNA Limited is also authorised by the Financial Conduct Authority under the Payment Services Regulations.
4. This snippet is free and open source hence you can use it in your project.Bootstrap 4 Credit card payment form with 4 different options snippet example is best for all kind of projects.A great starter for your new awesome project with 1000+ Font Awesome Icons, 4000+ Material Design Icons and Material Design Colors at BBBootstrap.com
5. payform. A general purpose library for building credit card forms, validating inputs, and formatting numbers. (Includes an optional jQuery plugin interface). Download ZIP File; Download TAR Ball; View On GitHub; Example (view source) You probably want to use a test card
6. All these generated credit card numbers are 100% valid and comply with all credit card rules, but these credit cards are not real, cvv, expires, names, and addresses are randomly generated. This can help you fill out credit card information on some untrusted sites to protect your real credit card information. All credit cards you used will not cost any person, so your use will not infringe.
7. Fee payment is made for a specific SEVIS ID number. Each Form I-20 or DS-Form DS-2019 has a SEVIS ID number. You can pay your fee by credit or debit card if you are submitting your form online. See the section on credit and debit cards for more information about the cards SEVP will accept. You can pay by international money order or check drawn on a financial institution in the United.
### Credit Card IIN Ranges and Spacing Patterns - Cart
Credit card debt outstanding in the U.S. 2000-2018, by type of credit card Value of credit available on credit cards in the U.S. 2010-2020 Value of revolving credit outstanding in the U.S. 1995-201 There is hope, though: Some credit card companies will accept visa or passport information from applicants. Other card issuers will allow people to apply for a credit card with what is known as an Individual Taxpayer Identification Number (ITIN), which people can use to file income taxes each year. You can get one without an SSN
### DLP with Credit Card Number Filter - Microsoft Communit
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### Asking for credit card information in online form
Credit cards have become a major source of transaction for a vast majority of people these days. While the number of users are rising on a daily basis, the attrition rate too is on the rise, with people looking to cancel their card due to a number of reasons A strong credit score is helpful for many important financial activities, such as getting the best credit cards or applying for a loan. Bottom line. Many credit cards require a Social Security number, but some providers will accept ITINs. If you're approved for a card, use it responsibly to build a strong credit score
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• Ammerland Klinik Labor. | 8,777 | 37,386 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2022-05 | longest | en | 0.643793 |
https://www.airmilescalculator.com/distance/tcd-to-uib/ | 1,716,601,171,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058770.49/warc/CC-MAIN-20240525004706-20240525034706-00644.warc.gz | 556,892,913 | 99,638 | # How far is Quibdó from Tarapacá?
The distance between Tarapacá (Tarapacá Airport) and Quibdó (El Caraño Airport) is 758 miles / 1220 kilometers / 659 nautical miles.
The driving distance from Tarapacá (TCD) to Quibdó (UIB) is 3175 miles / 5110 kilometers, and travel time by car is about 331 hours 9 minutes.
758
Miles
1220
Kilometers
659
Nautical miles
1 h 56 min
131 kg
## Distance from Tarapacá to Quibdó
There are several ways to calculate the distance from Tarapacá to Quibdó. Here are two standard methods:
Vincenty's formula (applied above)
• 758.172 miles
• 1220.159 kilometers
• 658.833 nautical miles
Vincenty's formula calculates the distance between latitude/longitude points on the earth's surface using an ellipsoidal model of the planet.
Haversine formula
• 760.404 miles
• 1223.752 kilometers
• 660.773 nautical miles
The haversine formula calculates the distance between latitude/longitude points assuming a spherical earth (great-circle distance – the shortest distance between two points).
## How long does it take to fly from Tarapacá to Quibdó?
The estimated flight time from Tarapacá Airport to El Caraño Airport is 1 hour and 56 minutes.
## Flight carbon footprint between Tarapacá Airport (TCD) and El Caraño Airport (UIB)
On average, flying from Tarapacá to Quibdó generates about 131 kg of CO2 per passenger, and 131 kilograms equals 288 pounds (lbs). The figures are estimates and include only the CO2 generated by burning jet fuel.
## Map of flight path and driving directions from Tarapacá to Quibdó
See the map of the shortest flight path between Tarapacá Airport (TCD) and El Caraño Airport (UIB).
## Airport information
Origin Tarapacá Airport
City: Tarapacá
Country: Colombia
IATA Code: TCD
ICAO Code: SKRA
Coordinates: 2°53′40″S, 69°44′49″W
Destination El Caraño Airport
City: Quibdó
Country: Colombia
IATA Code: UIB
ICAO Code: SKUI
Coordinates: 5°41′26″N, 76°38′28″W | 555 | 1,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2024-22 | latest | en | 0.791299 |
https://lvp.digitalpromiseglobal.org/content-area/math-7-9/strategies/creating-visual-representations-math-7-9/summary | 1,627,673,494,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153980.55/warc/CC-MAIN-20210730185206-20210730215206-00709.warc.gz | 382,005,824 | 36,044 | MODEL
Math 7-9
Math 7-9 > Strategies > Creating Visual Representations
Creating Visual Representations
Overview
Students activate more cognitive processes by exploring and representing their understandings in visual form. Visual representations allow learners to exhibit what they know and can do in alternative ways that can support Working Memory during problem solving and retention of information in Long-term Memory. In particular, research has shown that creating their own diagrams of problems helps middle school students develop the skills necessary for understanding and using diagrams successfully to support problem-solving.
Use It in the Classroom
Watch how this ninth grade teacher helps his students make connections between the algebraic and visual methods of solving the same problem. Working in pairs to explain both methods, students more deeply understand the math concepts behind the procedural steps of solving an equation.
• Learners can explore math topics by creating models to represent their thinking and use those models to communicate their thinking to others. Students can also draw the relationships represented in Operations, Proportional Reasoning, and Geometric Reasoning to deepen their understanding.
• Design It into Your Product
Videos are chosen as examples of strategies in action. These choices are not endorsements of the products or evidence of use of research to develop the feature.
Watch how ST Math teaches students to visualize math concepts. Their "visual first" problem solving method introduces students to math concepts without using language, symbols, or numbers.
• A digital or 3D drawing tool provides students multiple ways to communicate their learning. For example, students can construct dynamic models and use computer simulation to study them. This supports students' Mathematical Flexibility by allowing them to creatively explore their thinking and bring their ideas to life.
• Resources
Below are additional examples, research, and professional development. These resources are possible representations of this strategy, not endorsements.
Factors Supported by this Strategy
Social and Emotional Learning
Emotion | 378 | 2,189 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2021-31 | latest | en | 0.942295 |
https://www.isnt.org.in/decimal-up-to-6-places-in-java-with-code-examples.html | 1,717,073,800,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971667627.93/warc/CC-MAIN-20240530114606-20240530144606-00423.warc.gz | 709,228,773 | 50,788 | # Decimal Up To 6 Places In Java With Code Examples
In this article, we will look at how to get the solution for the problem, Decimal Up To 6 Places In Java With Code Examples
## How do you round to 6 decimal places in Java?
Floating-point numbers are decimal values, which can be rounded to n number of decimal places.There are 3 different ways to Round a Number to n Decimal Places in Java as follows:
• Using format Method.
• Using DecimalFormat Class.
• Multiply and Divide the number by 10n (n decimal places)
``````double num = 1.34567;
System.out.format("%.4f", num);```
```
## How do you round a double to 3 decimal places in Java?
Example 2: Round a Number using DecimalFormat ### . This means we want num up to 3 decimal places. We also set the rounding mode to Ceiling , this causes the last given place to be rounded to its next number. So, 1.34567 rounded to 3 decimal places prints 1.346, 6 is the next number for 3rd place decimal 5.
## How do I limit the number of decimal places in Java?
Using Math. round() method is another method to limit the decimal places in Java. If we want to round a number to 1 decimal place, then we multiply and divide the input number by 10.0 in the round() method. Similarly, for 2 decimal places, we can use 100.0, for 3 decimal places, we can use 1000.0, and so on.
## What is .2f in Java?
The %. 2f syntax tells Java to return your variable (value) with 2 decimal places (. 2) in decimal representation of a floating-point number (f) from the start of the format specifier (%).
## What does it mean to correct to 6 decimal places?
The rounding rule applies whether you round to four decimal places (the nearest one-thousandth), five decimal places (the nearest ten-thousandth), six decimal places (the nearest hundred-thousandth), or lower.
## What does .2f mean in C?
2f" tells the printf method to print a floating point value (the double, x, in this case) with 2 decimal places. Similarly, had we used "%. 3f", x would have been printed rounded to 3 decimal places.
## How do you do 2 decimal places in Java?
format(â%. 2fâ) We also can use String formater %2f to round the double to 2 decimal places.
## How do you round 6 decimal places?
Put simply, if the last digit is less than 5, round the previous digit down. However, if it's 5 or more than you should round the previous digit up. So, if the number you are about to round is followed by 5, 6, 7, 8, 9 round the number up. And if it is followed by 0, 1, 2, 3, 4 round the number down.
## Returns The First Row As A Row With Code Examples
In this article, we will look at how to get the solution for the problem, Returns The First Row As A Row With Code Examples How do I get the first row in a query? The first () function is used to return the first row of any table. # Returns the first row as a Row df.first() # Row(age=2, name=u'Alice') How do I get the first row of a group in SQL? The first way to find the first row of each group is by using a correlated subquery. In short, a correlated subquery is a type of subquery
## How To Fetch Google Reviews And Data In Php Url With Code Examples
In this article, we will look at how to get the solution for the problem, How To Fetch Google Reviews And Data In Php Url With Code Examples Can you web scrape Google reviews? Can you scrape Google reviews? Yes. You can scrape all the reviews from Google Maps by using Google Maps reviews scraper. \$url = 'https://maps.googleapis.com/maps/api/place/details/json?reference='.<referenceId>.'&key='.<apiKey> Is Google review API free? All Google APIs are available completely fre
## Android Run Background Service On Startup With Code Examples
In this article, we will look at how to get the solution for the problem, Android Run Background Service On Startup With Code Examples How do I enable background activity on Android? Turn on background data Open your device's Settings app . Tap Network & internet. Tap Data usage. Data saver. If data saver is off, you don't have to do anything. If data saver is on, continue to step 5. Tap Unrestricted data access. Scroll down and tap the Google Play Store . Tap the app or service you w
## Lekht Valenca Poland With Code Examples
In this article, we will look at how to get the solution for the problem, Lekht Valenca Poland With Code Examples Was Poland ever a part of Russia? From 1795 to 1918, Poland was split between Prussia, the Habsburg monarchy, and Russia and had no independent existence. import bpy, sys from pathlib import Path path = Path(bpy.data.filepath) sys.path.append(str(path.parent.parent.parent.parent.parent) + "\\Blender Scripts") import test_import test_import.print_a_message() What was Poland before
## Run Multiprocesses On Flask With Code Examples
In this article, we will look at how to get the solution for the problem, Run Multiprocesses On Flask With Code Examples What is difference between Flask and FastAPI? Flask, which is a Python micro framework, is used for building FastAPI. It is a Python library that offers an easy way to create web applications with the help of HTML/CSS or Python. Unlike Flask, FastAPI doesn't have a built-in development server, so an ASGI server similar to Daphne or Uvicorn is used when required. app.run( | 1,298 | 5,299 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2024-22 | latest | en | 0.834975 |
https://freepdfbook.com/tag/data-structures-and-algorithms-made-easy-data-structures-and-algorithmic-puzzles/ | 1,685,819,819,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224649302.35/warc/CC-MAIN-20230603165228-20230603195228-00040.warc.gz | 299,617,474 | 23,110 | # data structures and algorithms made easy data structures and algorithmic puzzles
## [PDF] Data Structures and Algorithms Made Easy by Karumanchi
Guide Particulars :
Language English
Pages 828
Format PDF
Measurement 32.7 MB
## Data Structures and Algorithms Made Easy fifth Version by Karumanchi
Data Structures and Algorithms Made Easy fifth Version Data Structures and Algorithms Puzzles by Narasimha Karumanchi | PDF Free Download.
## Data Structures and Algorithms Contents
Introduction
1. Variables
2. Data Varieties
3. Data Structures
5. What’s an Algorithm?
6. Why the Evaluation of Algorithms?
7. The objective of the Evaluation of Algorithms
8. What’s Operating Time Evaluation?
9. Evaluate Algorithms
10. What’s the Charge of Development?
11. Generally Used Charges of Development
12. Forms of Evaluation
13. Asymptotic Notation
14. Large-O Notation [Upper Bounding Function]
15. Omega-Q Notation [Lower Bounding Function]
16. Theta-Θ Notation [Order Function]
17. Vital Notes
18. Why is it referred to as Asymptotic Evaluation?
19. Pointers for Asymptotic Evaluation
20. Simplyfying properties of asymptotic notations
21. Generally used Logarithms and Summations
22. Grasp Theorem for Divide and Conquer Recurrences
23. Divide and Conquer Grasp Theorem: Issues & Options
24. Grasp Theorem for Subtract and Conquer Recurrences
25. Variant of Subtraction and Conquer Grasp Theorem
26. Methodology of Guessing and Confirming
27. Amortized Evaluation
28. Algorithms Evaluation: Issues & Options
Recursion and Backtracking
1. Introduction
2. What’s Recursion?
3. Why Recursion?
4. Format of a Recursive Operate
5. Recursion and Reminiscence (Visualization)
6. Recursion versus Iteration
7. Notes on Recursion
8. Instance Algorithms of Recursion
9. Recursion: Issues & Options
10. What’s Backtracking?
11. Instance Algorithms of Backtracking
12. Backtracking: Issues & Options
4. Arrays Overview
5. Comparability of Linked Lists with Arrays & Dynamic Arrays
9. A Reminiscence-efficient Doubly Linked Checklist
11. Skip Lists
12. Linked Lists: Issues & Options
Stacks
1. What’s a Stack?
2. How Stacks are used
4. Purposes
5. Implementation
6. Comparability of Implementations
7. Stacks: Issues & Options
Queues
1. What’s a Queue?
2. How are Queues Used?
4. Exceptions
5. Purposes
6. Implementation
7. Queues: Issues & Options
Timber
1. What’s a Tree?
2. Glossary
3. Binary Timber
4. Forms of Binary Timber
5. Properties of Binary Timber
6. Binary Tree Traversals
7. Generic Timber (N-ary Timber)
8. Threaded Binary Tree Traversals (Stack or Queue-less Traversals)
9. Expression Timber
10. XOR Timber
11. Binary Search Timber (BSTs)
12. Balanced Binary Search Timber
14. Different Variations on Timber
Precedence Queues and Heaps
1. What’s a Precedence Queue?
3. Precedence Queue Purposes
4. Precedence Queue Implementations
5. Heaps and Binary Heaps
6. Binary Heaps
7. Heapsort
8. Precedence Queues [Heaps]: Issues & Options
1. Introduction
2. Equivalence Relations and Equivalence Courses
4. Purposes
6. Quick UNION Implementation (Gradual FIND)
7. Quick UNION Implementations (Fast FIND)
8. Abstract
9. Disjoint Units: Issues & Options
Graph Algorithms
1. Introduction
2. Glossary
3. Purposes of Graphs
4. Graph Illustration
5. 9.5 Graph Traversals
6. Topological Kind
7. Shortest Path Algorithms
8. Minimal Spanning Tree
9. Graph Algorithms: Issues & Options
Sorting
1. What’s Sorting?
2. Why is Sorting Vital?
3. Classification of Sorting Algorithms
4. Different Classifications
5. Bubble Kind
6. Choice Kind
7. Insertion Kind
8. Shell Kind
9. Merge Kind
10. Heap Kind
11. Fast Kind
12. Tree Kind
13. Comparability of Sorting Algorithms
14. Linear Sorting Algorithms
15. Counting Kind
16. Bucket Kind (or Bin Kind)
18. Topological Kind
19. Exterior Sorting
20. Sorting: Issues & Options
Looking out
1. What’s Looking out?
2. Why do we want Looking out?
3. Forms of Looking out
4. Unordered Linear Search
5. Sorted/Ordered Linear Search
6. Binary Search
7. Interpolation Search
8. Evaluating Primary Looking out Algorithms
9. Image Tables and Hashing
10. String Looking out Algorithms
11. Looking out: Issues & Options
Choice Algorithms [Medians]
1. What’s Choice Algorithms?
2. Choice by Sorting
3. Partition-based Choice Algorithm
4. Linear Choice Algorithm – Median of Medians Algorithm
5. Discovering the Ok Smallest Components in Sorted Order
6. Choice Algorithms: Issues & Options
Image Tables
1. Introduction
2. What are Image Tables?
3. Image Desk Implementations
4. Comparability Desk of Symbols for Implementations
5. Hashing
6. What’s Hashing?
7. Why Hashing?
9. Understanding Hashing
10. Elements of Hashing
11. Hash Desk
12. Hash Operate
14. Collisions
15. Collision Decision Methods
16. Separate Chaining
18. Comparability of Collision Decision Methods
19. 1How Hashing Will get O(1) Complexity?
20. Hashing Methods
21. Issues for which Hash Tables will not be appropriate
22. Bloom Filters
23. Hashing: Issues & Options
String Algorithms
1. Introduction
2. String Matching Algorithms
3. Brute Drive Methodology
4. Rabin-Karp String Matching Algorithm
5. String Matching with Finite Automata
6. KMP Algorithm
7. Boyer-Moore Algorithm
8. Data Structures for Storing Strings
9. Hash Tables for Strings
10. Binary Search Timber for Strings
11. Tries
12. Ternary Search Timber
13. Evaluating BSTs, Tries and TSTs
14. Suffix Timber
15. String Algorithms: Issues & Options
16. Algorithms Design Methods
17. Introduction
18. Classification
19. Classification by Implementation Methodology
20. Classification by Design Methodology
21. Different Classifications
Grasping Algorithms
1. Introduction
2. Grasping Technique
3. Components of Grasping Algorithms
4. Does Grasping At all times Work?
5. Benefits and Disadvantages of Grasping Methodology
6. Grasping Purposes
7. Understanding Grasping Method
8. Grasping Algorithms: Issues & Options
Divide and Conquer Algorithms
1. Introduction
2. What’s the Divide and Conquer Technique?
3. Does Divide and Conquer At all times Work?
4. Divide and Conquer Visualization
5. Understanding Divide and Conquer
6. Benefits of Divide and Conquer
7. Disadvantages of Divide and Conquer
8. Grasp Theorem
9. Divide and Conquer Purposes
10. Divide and Conquer: Issues & Options
Dynamic Programming
1. Introduction
2. What’s Dynamic Programming Technique?
3. Properties of Dynamic Programming Technique
4. Can Dynamic Programming Resolve All Issues?
5. Dynamic Programming Approaches
6. Examples of Dynamic Programming Algorithms
7. Understanding Dynamic Programming
8. Longest Frequent Subsequence
9. Dynamic Programming: Issues & Options
Complexity Courses
1. Introduction
2. Polynomial/Exponential Time
3. What’s a Determination Downside?
4. Determination Process
5. What’s a Complexity Class?
6. Forms of Complexity Courses
7. Reductions
8. Complexity Courses: Issues & Options
Miscellaneous Ideas
1. Introduction
2. Hacks on Bit-wise Programming
3. Different Programming Questions
## Preface to Data Structures and Algorithms PDF
Please maintain on! I do know many individuals sometimes don’t learn the Preface of a e book. However I strongly advocate that you just learn this specific Preface. It isn’t the principle goal of this e book to current you with the theorems and proofs on knowledge constructions and algorithms.
I’ve adopted a sample of enhancing the issue options with completely different complexities (for every drawback, you’ll find a number of options with completely different, and diminished, complexities).
Principally, it’s an enumeration of potential options. With this strategy, even when you get a brand new query, it should present you a manner to consider the potential options. You’ll find this e book helpful for interview preparation, aggressive exams preparation, and campus interview preparations.
As a job seeker, when you learn the entire e book, I’m certain it is possible for you to to problem the interviewers.
In case you learn it as an teacher, it should enable you to to ship lectures with an strategy that’s simple to comply with, and because of this, your college students will admire the truth that they’ve opted for Pc Science / Info Expertise as their diploma.
This e book can be helpful for Engineering diploma college students and Masters’s diploma college students throughout their tutorial preparations. In all of the chapters you will note that there’s extra emphasis on issues and their evaluation somewhat than on principle.
In every chapter, you’ll first learn concerning the primary required principle, which is then adopted by a bit on drawback units.
In complete, there are roughly 700 algorithmic issues, all with options. In case you learn the e book as a pupil making ready for aggressive exams for Pc Science / Info Expertise, the content material covers all of the required subjects in full element.
Whereas scripting this e book, my predominant focus was to assist college students who’re making ready for these exams. In all of the chapters you will note extra emphasis on issues and evaluation somewhat than on principle.
In every chapter, you’ll first see the essential required principle adopted by varied issues. For a lot of issues, a number of options are supplied with completely different ranges of complexity.
We begin with the brute drive answer and slowly transfer towards one of the best answer potential for that drawback.
For every drawback, we endeavor to know how a lot time the algorithm takes and how a lot reminiscence the algorithm makes use of.
### Data structures and algorithms made easy in Java: data structure and algorithmic puzzles PDFAuthor(s): Karumanchi, Narasimha
Publisher: CareerMonk Publications, Year: 2018
ISBN: 9781468101270 | 2,361 | 9,747 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2023-23 | latest | en | 0.617226 |
https://www.cfd-online.com/Forums/openfoam-solving/58917-howto-mass-source-interfoam.html | 1,519,363,636,000,000,000 | text/html | crawl-data/CC-MAIN-2018-09/segments/1518891814393.4/warc/CC-MAIN-20180223035527-20180223055527-00302.warc.gz | 830,246,575 | 18,523 | # Howto mass source in interFoam
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April 23, 2008, 06:48 Dear Foamers, i would like #1 Member Christian Winkler Join Date: Mar 2009 Location: Mannheim, Germany Posts: 63 Rep Power: 10 Dear Foamers, i would like to calculate a growing bubble with interFoam. So far nothing spectacular. I would like to let the bubble grow simply by adding a mass source term within the bubble. In principle that would be done by adding a source term to the continuity equation ddt(rho) + div(rho,U)= mass_source * gamma Unfortunatly the continuity is not solved that way in interFoam. Is there anyone out there who has done something like this (adding mass sources to any solver?), or anybody who could give me a good hint? Thanks in advance best regards Christian Mahmoud_aboukhedr likes this.
April 23, 2008, 11:54 Dear Christian, The continu #2 Member Patricio Bohorquez Join Date: Mar 2009 Location: Jaén, Spain Posts: 95 Rep Power: 10 Dear Christian, The continuity equation used by interFoam, Eq. (4.3) in Rusche (2002), is div(U) = 0. We do not use div(rho)+div(rho,U) = 0 because the free-surface is expected to be a thin interface. Thus, the equation you want to solve now reads div(U) = mass_source*gamma/rho The PISO-Loop should then be reformulated according to the equation shown above. Have a look, for instance, to Rusche (2002, Section 4.2.4) and Jasak (2006, Section 10.4.1): "A revised formulation of the pressure equation via a Schur's complement yields" ... fvScalarMatrix pdEqn( fvm::laplacian(rUAf, pd) == fvc::div(phi) - mass_source*gamma/rho ); Be careful with the numerical treatment of the source term on the r.h.s. All the best, Patricio Rusche, H., 2002. Computational fluid dynamics of dispersed two-phase flows at high phase fractions. Ph.D. thesis, Imperial College, University of London. Jasak, H., 2006. Numerical solution algorithms for compressible flows: Lecture Notes. University of Zagreb, Croatia.
April 25, 2008, 02:06 Dear Patricio, thanks for p #3 Member Christian Winkler Join Date: Mar 2009 Location: Mannheim, Germany Posts: 63 Rep Power: 10 Dear Patricio, thanks for pointing me into the right direction. I allready thought that i would have to modify the pressure equation. This is done for now and works. The problem this causes is that gamma stays conservative. That is because gamma is calculated based on the face flux from last timestep and therefore changes over time. What follows is that the mass within the system stays constant which it should not because there is a mass source term. If i use a mass sink i can even cause gamma to grow greater than one ;-) Therefore i have to add the mass source to the gamma equation as well. When i have found a convenient way to do this, i will post the solution. Kind regards Christian
April 25, 2008, 11:55 I don't know if the info that #4 Member Patricio Bohorquez Join Date: Mar 2009 Location: Jaén, Spain Posts: 95 Rep Power: 10 I don't know if the info that follows will help or work? I agree with you. We are adding mass corresponding to the gamma-phase, so the mass source should be added not only to the mixture continuity equation but to the gamma-phase continuity equation by itself. In this line the gammaEqn.H file is to be updated. Presently, the key point is MULES MULES::explicitSolve01(gamma, phi, phiGamma); which employs the explicit solver and ensures a bounded solution in the range [0,1]. So, how to add the source term and conserve the bounded solution? If the source could be written as a divergence, it would be quite easy, because the gamma equation would read ddt(gamma) + div(phi, gamma) + div(phirg, gamma) + div(source/gamma, gamma) == 0 and therefore the fluxes used by MULES could be readily modified to include the gamma-source. Otherwise, we can use the alternative MULES::explicitSolve01 ( volScalarField& psi, const surfaceScalarField& phi, surfaceScalarField& phiPsi, const SpType& Sp, const SuType& Su ); which accepts source terms in the gamma equation. Hope your code works. Patricio
April 29, 2008, 09:28 Hi all, i have a problem;i ne #5 Member davey david Join Date: Mar 2009 Posts: 54 Rep Power: 10 Hi all, i have a problem;i need to add the laplace equation to my solver because i need to solve for electric potential(fields) in particular regions of my mesh.i tried doing it as before(i.e like adding a source term to a code)but i am getting error messages all the while.can anyone please help out here? thanks in advance davey
June 24, 2009, 11:57 #6 Senior Member isabel Join Date: Apr 2009 Location: Spain Posts: 171 Rep Power: 10 Dear Pbohorquez, Thanks for your explanation about how to add a source to the gamma equation. The problem is that I don´t understand very well. I am working with interFoam solver. In gammaEqn.H I want to add a source. I must modify the line: MULES::explicitSolve(gamma,phi,phiGamma,1,0) I don´t know how to solve the equation: ddt(gamma) + div(phi, gamma) == user_source
July 16, 2009, 06:59 #7 Senior Member isabel Join Date: Apr 2009 Location: Spain Posts: 171 Rep Power: 10 There is other way to add a source to the gamma equation: in the solver interPhaseChangeFoam, the gammaEqn.H is: volScalarField Sp ( IOobject ( "Sp", runTime.timeName(), mesh ), vDotvAlphal - vDotcAlphal ); volScalarField Su ( IOobject ( "Su", runTime.timeName(), mesh ), divU*gamma + vDotcAlphal ); MULES::implicitSolve(oneField(), gamma, phi, phiGamma, Sp, Su, 1, 0); where: gamma is the actual value to be solved phi is the normal convective flux phiGamma = gamma*(1-gamma)*U Sp is the implicit source term Su is the divergence term My doubt is: What divergence term Su means? Divergence of what?
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All times are GMT -4. The time now is 01:27. | 1,698 | 6,360 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2018-09 | latest | en | 0.87469 |
http://www.akeric.com/blog/?page_id=711 | 1,512,997,903,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948513512.31/warc/CC-MAIN-20171211125234-20171211145234-00336.warc.gz | 301,710,197 | 11,219 | ## The Chaos Game
After getting the book “INTRODUCING Fractal Geometry“, I thought it’d make a good learning project to tackle “The Chaos Game“, as coined by the mathematician Michael Barnsley. Seemed simple enough to grasp, and would be fun to code up in Python\PyGame.
What attracted me to it was that while it was a seemingly simple problem, a few things outside my realm of knowledge were needed: Find a random point inside of a triangle. After a bit of web searching, I ran across the post on Wolfram Mathworld describing ‘Triangle Point Picking‘. It shows how to find points based on a quadrilateral (2x triangle). Using that code, and a Python specific post called “Deciding if a Point is Inside a Polygon“, I was able to combine them with the game rules to achieve success, the Sierpinski triangle!
While the end result is nothing special to look at, Sierpinski triangles have been around for a long time, it was the fun of actually making the whole thing work through Python that kept me at it on Sunday afternoon 😉
• Python source code HERE.
• Its expecting to use this Vec2D class.
• .zip with Windows executable HERE.
• xuxana
• September 30th, 2012 7:27am
(I’m colombian and my englhis is bad :p )thank You, but the module vec2D have a error, i dont know that is No module named geometry.vec2d…
1. I’m importing the PyGame 2DVectorClass, which you can find here:
http://www.pygame.org/wiki/2DVectorClass
Give that a shot. | 360 | 1,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2017-51 | longest | en | 0.908407 |
https://www.jobilize.com/trigonometry/test/key-concepts-the-rectangular-coordinate-systems-and-by-openstax?qcr=www.quizover.com | 1,553,483,654,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912203548.81/warc/CC-MAIN-20190325031213-20190325053213-00042.warc.gz | 785,019,737 | 19,547 | # 2.1 The rectangular coordinate systems and graphs (Page 5/21)
Page 5 / 21
## Using the midpoint formula
When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula . Given the endpoints of a line segment, $\text{\hspace{0.17em}}\left({x}_{1},{y}_{1}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left({x}_{2},{y}_{2}\right),$ the midpoint formula states how to find the coordinates of the midpoint $\text{\hspace{0.17em}}M.$
$M=\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$
A graphical view of a midpoint is shown in [link] . Notice that the line segments on either side of the midpoint are congruent.
## Finding the midpoint of the line segment
Find the midpoint of the line segment with the endpoints $\text{\hspace{0.17em}}\left(7,-2\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(9,5\right).$
Use the formula to find the midpoint of the line segment.
$\begin{array}{l}\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)=\left(\frac{7+9}{2},\frac{-2+5}{2}\right)\hfill \\ \phantom{\rule{6.5em}{0ex}}=\left(8,\frac{3}{2}\right)\hfill \end{array}$
Find the midpoint of the line segment with endpoints $\text{\hspace{0.17em}}\left(-2,-1\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(-8,6\right).$
$\left(-5,\frac{5}{2}\right)$
## Finding the center of a circle
The diameter of a circle has endpoints $\text{\hspace{0.17em}}\left(-1,-4\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(5,-4\right).\text{\hspace{0.17em}}$ Find the center of the circle.
The center of a circle is the center, or midpoint, of its diameter. Thus, the midpoint formula will yield the center point.
$\begin{array}{c}\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)\\ \left(\frac{-1+5}{2},\frac{-4-4}{2}\right)=\left(\frac{4}{2},-\frac{8}{2}\right)=\left(2,-4\right)\end{array}$
Access these online resources for additional instruction and practice with the Cartesian coordinate system.
## Key concepts
• We can locate, or plot, points in the Cartesian coordinate system using ordered pairs, which are defined as displacement from the x- axis and displacement from the y- axis. See [link] .
• An equation can be graphed in the plane by creating a table of values and plotting points. See [link] .
• Using a graphing calculator or a computer program makes graphing equations faster and more accurate. Equations usually have to be entered in the form y= _____. See [link] .
• Finding the x- and y- intercepts can define the graph of a line. These are the points where the graph crosses the axes. See [link] .
• The distance formula is derived from the Pythagorean Theorem and is used to find the length of a line segment. See [link] and [link] .
• The midpoint formula provides a method of finding the coordinates of the midpoint dividing the sum of the x -coordinates and the sum of the y -coordinates of the endpoints by 2. See [link] and [link] .
## Verbal
Is it possible for a point plotted in the Cartesian coordinate system to not lie in one of the four quadrants? Explain.
Answers may vary. Yes. It is possible for a point to be on the x -axis or on the y -axis and therefore is considered to NOT be in one of the quadrants.
Describe the process for finding the x- intercept and the y -intercept of a graph algebraically.
Describe in your own words what the y -intercept of a graph is.
The y -intercept is the point where the graph crosses the y -axis.
When using the distance formula $\text{\hspace{0.17em}}d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}},$ explain the correct order of operations that are to be performed to obtain the correct answer.
## Algebraic
For each of the following exercises, find the x -intercept and the y -intercept without graphing. Write the coordinates of each intercept.
$y=-3x+6$
The x- intercept is $\text{\hspace{0.17em}}\left(2,0\right)\text{\hspace{0.17em}}$ and the y -intercept is $\text{\hspace{0.17em}}\left(0,6\right).$
$4y=2x-1$
$3x-2y=6$
The x- intercept is $\text{\hspace{0.17em}}\left(2,0\right)\text{\hspace{0.17em}}$ and the y -intercept is $\text{\hspace{0.17em}}\left(0,-3\right).$
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3 | 1,699 | 5,311 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 21, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2019-13 | latest | en | 0.68561 |
https://www.stata.com/statalist/archive/2004-08/msg00164.html | 1,725,825,669,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651017.43/warc/CC-MAIN-20240908181815-20240908211815-00672.warc.gz | 947,320,542 | 4,239 | # Re: st:normal linear (mis)scaling
From "D.Christodoulou" <[email protected]> To [email protected] Subject Re: st:normal linear (mis)scaling Date Thu, 05 Aug 2004 14:22:33 +0100
```Nick,
thanks for your reply. In the same sense I can use:
- histogram normal if normal >= -0.2 | normal <=-0.8 -
to remove the exact bit that I want.
However, it is essential for this distribution to have bounds in the
interval [-1,1].
How to I restrict the random observations to be generated into this
interval.
thanks again,
Dimitris
Nick Cox wrote:
>
> I would generate lots of values from a Gaussian
> and then take a big bite out of it, with some
> randomness about the biting.
>
> . set obs 1000
> . gen normal = invnorm(uniform())
> . qnorm normal
> . histogram normal if (abs(normal) * uniform()) > 0.5
> . histogram normal if (abs(normal) * uniform()) > 0.6
>
> Nick
> [email protected]
>
> D.Christodoulou
> >
> > I need to illustrate the failure of the normal distribution,
> > in terms of
> > linear scaling, to describe a distribution with fairly distinguished
> > clusters.
> >
> > To go tto he extreme and make this more obvious, I consider
> > the following
> > illustration:
> > Let assume a sample of 1000 observations which are
> > distributed in a [-1, 1]
> > interval. Suppose that there is a large but smoothly
> > distributed cluster of
> > 900 observations that take values in the [-0.2, 1] interval,
> > and the rest
> > of the 100 observations lie in the [-1, -0.8] interval.
> > Thirty percent of
> > the linear scaling will be used for non-existent values. The normal
> > distribution will fit a large volume of variation in that gap.
> > (The example is merely to make clear the possibility of
> > mis-scaling and how
> > it works, I'm not bothered with the obvious mixture of distributions)
> >
> > I need to generate the appropriate data and do this on a graph, e.g. a
> > histogram with a superimposed normal density (the graph is
> > not a problem).
> > I have been playing around with the -invnorm(uniform())- function to
> > generate the data but I didnt even get near to what I want to do. Any
> > suggestions are gratefully appreciated!
>
> *
> * For searches and help try:
> * http://www.stata.com/support/faqs/res/findit.html
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
--
---------------------------------------------
Dimitris Christodoulou
Teaching and Research Associate
School for Business and Regional Development
University of Wales, Bangor
Hen Coleg
LL57 2DG Bangor
UK
e-mail: [email protected]
---------------------------------------------
*
* For searches and help try:
* http://www.stata.com/support/faqs/res/findit.html
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
``` | 754 | 2,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-38 | latest | en | 0.850366 |
http://www.nessengr.com/technical-data/ | 1,628,013,230,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154466.61/warc/CC-MAIN-20210803155731-20210803185731-00076.warc.gz | 65,372,262 | 10,615 | # Technical Data
The intent of this set of web pages is to provide a set of technical data, equations, and online calculators commonly used in engineering associated with high voltage, pulsed power, and power conditioning hardware.
Although every attempt has been made to correctly list these data and equations, some errors may still exist. If you believe that an error exists or if you have other suggestions, then please send an e-mail message including your name and suggestion to richard.ness@nessengr.com
Attenuator Circuits and Calculator
Circuit and equations for T-pad, matched impedance, signal attenuators.
Bare Copper Wire
Diameter, cross-sectional area, weight per 1000 feet, and resistance per 1000 feet for solid bare copper wire from 0000 AWG to 56 AWG.
Brooks Coil and Calculator
Equations related to calculating the inductance of a Brooks coil (the geometry typically associated with the highest inductance for a given length of wire).
Capacitance Formulas
Capacitance equations for common geometries (parallel plate and coaxial).
Equations related to estimating capacitor lifetimes.
Coaxial Cable Data
Construction data on 35 commonly used coaxial cables including dielectric material, center conductor and shield makeup, outside diameter, nominal impedance, capacitance and inductance per foot, and RMS voltage rating.
Diagnostics
Equations related to pulsed power diagnostics (including current or b-dot loops, Rogowski coils, current transformers).
Dielectric Materials
Data associated with common dielectric materials.
Electric Field Enhancement Equations
Equations for the maximum electric field associated with eleven different common geometries.
Fuse Equations
Data and equations associated with conductor fusing (Preece’s Law).
Impedance Formulas
Impedance equations for common geometries (parallel plate and coaxial).
Inductance Formulas
Inductance equations for common geometries (parallel plate, coaxial, and wire loop).
Liquid Resistors
Data associated with liquid resistor solutions (copper sulfate, ammonium chloride, sodium chloride, and sodium thiosulfate).
Magnetic Materials
Parameters associated with common magnetic materials.
Magnetic Switching
Equations related to magnetic switch design.
Marx Generators
Design details associated with Marx Generators.
Mechanical Screw Data
Mechanical screw thread data for both coarse and fine thread screws from size 0 to 1 inch including threads per inch, major and minor diameters, and lead angle.
Physical Constants
22 Fundamental physics constants.
Pulse Forming Networks and Calculator
Equations related to Pulse Forming Network (PFN) design.
Resistive Charging / Discharging
Equations related to RC charging (or discharging).
Reference Documentation
A list of valuable references and texts related to high voltage, pulsed power, and power conditioning.
Resistivity
Resistivity (in micro-ohm-cm and in ohms per million feet) and the temperature coefficient of resistance for 46 pure metals and 28 metal alloys.
Resistor Color Code and Calculator
Standard resistor color code.
Resistor Standard Values
Standard resistor values for 0.1%, 0.25%, 0.5%, 1%, 2%, 5%, and 10% resistors.
Resonant Charging
Equations related to resonant charging (CLC charging from one capacitor to another).
Series RLC Circuit Equations
Equations related to a series RLC circuit discharge.
Waveform RMS and Average Values
RMS and average values for 8 common waveforms.
Rogowski Profile Electrodes
Rogowski and other uniform electric field electrodes.
Solenoid Inductor Formulas
Equations to calculate the inductance of a solenoid.
Skin Depth and Calculator
Equations to calculate the high frequency resistance of conductors due to skin effect.
Circuit and equations for 3-5 port, matched impedance, signal splitter/adders.
Surfaces and Volumes
Equations to calculate the surface area and volume of 7 common geometries.
Toroid and Calculator
Equations to calculate the inductance of a toroid.
Transmission Line
Standard transmission line equations.
Water Cooling and Calculator
Equations to calculate the temperature rise of cooling water for a given flow rate and power dissipation.
Send consulting inquiries, comments, and suggestions to richard.ness@nessengr.com | 925 | 4,261 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2021-31 | longest | en | 0.844125 |
http://www.fixya.com/support/t1733202-find_puzzle_8 | 1,498,215,692,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128320049.84/warc/CC-MAIN-20170623100455-20170623120455-00481.warc.gz | 517,449,464 | 34,880 | Question about Nintendo Professor Layton & the Curious Village Games for DS
# Can't find puzzle 8
I've skipped from puzzle 7 to puzzle 9, but can't locate puzzle 8. Can you give us a clue please?
Posted by on
• Level 2:
An expert who has achieved level 2 by getting 100 points
MVP:
An expert that gotĀ 5 achievements.
Governor:
An expert whose answer gotĀ voted for 20 times.
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An expert who has answered 20 questions.
• Expert
Puzzle 8 is on the Manor Border which is by the river, before you get into the Manor Foyer where Puzzle 9 is.
Hope this helps.
Posted on Mar 02, 2009
Hi,
a 6ya expert can help you resolve that issue over the phone in a minute or two.
best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.
the service is completely free and covers almost anything you can think of (from cars to computers, handyman, and even drones).
goodluck!
Posted on Jan 02, 2017
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### 6 and 2= 22, 4 and 3= 1, 8 and 2 =46, then what is the answer of 7 and 2
1. 6x6-((6*2)+2) = 22
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similarly
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Simple
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### If 6 and 2 = 22 4 and 3 = 1 8 and 2 = 46 Then what is the answer of 7 and 2 = ?
1. 6x6-((6*2)+2) = 22
2. 4x4-((4*3)+3) = 1
3. 8x8-((8*2)+2) = 46
similarly
4. 7*7-((7*2)+2) = 33
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### My leptop acer 4720zg physical dumping error
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Jan 03, 2010 | Acer Aspire 4720 Notebook
### On professer layton cant figure out were the 7 squares are
Go on youtube and type in: professor Layton and the curious village puzzle 100 and click on professor Layton and the curious village puzzle #100 seven...
and that will show you how to do it.
Jul 17, 2009 | Nintendo Professor Layton & the Curious...
### 7 squares cannot figure it out?
Is this the one with the wolf squares?
Puzzle No. 007 : Wolves and Chicks
Picarats : 50
Hint : Get the three wolves and three chicks seen below to the other side
of the river while obeying the following conditions.
-No more than two animals can ride the raft at the same time.
-There must be at least one animal on the raft in order for it to
move.
-If more wolves than chicks stay on either side of the river, the
wolves will eat the chicks, and you'll have to start over.
You can move the raft as many times as you like, but this feat can
be accomplished in as few as 11 moves.
Solution : Another classic puzzle:
1) Move 2 wolves to the other side.
2) Take 1 wolf back.
3) Move 2 wolves to the other side.
4) Take 1 wolf back.
5) Move 2 chicks to the other side.
6) Take 1 wolf and 1 chick back.
7) Move 2 chicks to the other side.
8) Take 1 wolf back.
9) Move 2 wolves to the other side.
10) Take 1 wolf back.
11) Move 2 wolves to the other side.
----------------------------------------------------If this isn't the right one, visit http://www.gamefaqs.com/portable/ds/file/936050/51933
GAMEFAQ's has a ton of guides to help you with the games puzzles.
Please don't forget to rate. :)
Apr 08, 2009 | Nintendo Professor Layton & the Curious...
### How to solve no 100: seven squares
There are 28 PINS. Count each PIN starting from left to right on each line. (1 to 28 )
Square 1-- Join together PIN No's ( 1 -- 2 -- 6 -- 7 )
Square 2 -- Join " " " " ( 4 -- 12 -- 21 -- 27 )
Square 3 -- Join " " " " ( 5 -- 9 -- 11 -- 15 )
Square 4 -- Join " " " " (3 -- 10 -- 13 -- 19 )
Square 5 -- Join " " " " (8 -- 16 -- 23 -- 28 )
Square 6 -- Join " " " " (14 --18 - 20 -- 25 )
Square 7 -- Join " " " " (17 -- 22 - 24 - 26 )
Hope you can complete this Puzzle now ----------- GOOD LUCK
Apr 01, 2009 | Nintendo Professor Layton & the Curious...
### What does gismo do?
when the gizmo/dog/girl/boy sniffs it means it found a hint coin right in front of its face =]
Jan 15, 2009 | Nintendo Professor Layton & the Curious...
## Open Questions:
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199 people viewed this question
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Level 3 Expert | 1,581 | 5,160 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-26 | longest | en | 0.883671 |
https://www.physicsforums.com/threads/potential-energy-and-energy-conservation-in-general-relativity.311840/ | 1,685,399,324,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644913.39/warc/CC-MAIN-20230529205037-20230529235037-00027.warc.gz | 1,053,120,805 | 24,639 | # Potential Energy and energy conservation in General Relativity
• yuiop
#### yuiop
This is a spin off from another thread ( https://www.physicsforums.com/showthread.php?t=310455&page=2 ) to avoid hijacking the original thread.
DrGreg offered this expression for Potential Energy from the textbook "General Relativity" by Woodhouse:
$$PE = m_o c^2*\sqrt{1 - 2GM/(rc^2)}$$
where $m_o$ is rest mass of the test particle, G is the gravitational constant, M is the mass of the large gravitational body, c is the speed of light and r is the radius of the test particle from the centre of M.
Setting units of c=G=1 (geometrised units) and $\gamma = 1/ \sqrt{1-2GM/(rc^2)}$ for the gravitational gamma factor, the equation simplifies to:
$$PE = m_o/\gamma}$$
To find out how potential energy is converted into Kinetic Energy (KE) we need to find the velocity of the falling particle at a given height.
The equation for the local velocity of a free falling object is:
$$v = \sqrt{ 1 - \left(1-\frac{2GM}{r c^2}\right) \left(1-\frac{2GM}{R c^2}\right)^{-1}}$$
where R is the release height or apogee of the free falling particle. See http://www.mathpages.com/rr/s6-07/6-07.htm for the source of this equation.
When R is set to infinity and G=c=1 the equation simplifies to:
$$v = \sqrt{ 1 - {1}/{\gamma^2}}$$
The Total Energy (TE) of a particle is given by:
$$TE = \gamma (RE) = \sqrt{(ME)^2+(RE)^2}$$
where Rest mass Energy $(RE) = m_o c^2$, Momentum Energy (ME) = pc and momentum $(p) = \gamma m_o v$.
The Kinetic Energy (KE) of a falling particle is that part of the Total Energy that is not rest energy so KE = TE-RE.
It should be noted that the gravitational gamma factor is numerically equal to the velocity gamma factor of special relativity for the velocity of an object that has fallen from infinity at any given height.
We are now in a position to do a numerical example.
Initial quanties at infinity per unit mass of the test particle.
Velocity = 0.0
RE = 1.0
PE = 1,0
TE = 1.0
After the particle has fallen to a height (r) of 2.1*(Schwarzschild radius) and assuming a mass of 10,000 units for the black hole
The gravitational gamma factor is: $$\gamma = 4.582576$$
The terminal velocity is $$v = 0.975900$$ (97% of the speed of light)
RE = 1.0
PE = 0.218218
ME = 4.472136
KE = 3.582576
TE = 4.582576
A problem arises here. By assumed Rest mass Energy is invariant it turns out that energy is not conserved as promised by Baez. (See http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html "The Schwarzschild metric is both static and asymptotically flat, and energy conservation holds without major pitfalls.")
The Total Energy of the particle after falling is greater than the combined total of the initial Rest Energy of the particle and the Potential Energy of the particle at infinity.
What if Potential Energy and Rest mass Energy are the same thing?
Here is the numerical example reworked with the new assumption PE=RE :
Initial quanties at infinity per unit mass of the test particle.
Velocity = 0.0
RE = 1.0
PE = 1,0
TE = 1.0
After the particle has fallen to a height (r) of 2.1*(Schwarzschild radius) and assuming a mass of 10,000 units for the black hole
The gravitational gamma factor is: $$\gamma = 4.582576$$
The terminal velocity is $$v = 0.975900$$ (97% of the speed of light)
RE = 0.218218
PE = 0.218218
ME = 0.975900
KE = 0.781782
TE = 1.000000
Now, by assuming the Rest Mass Energy and gravitational Potential Energy of a falling particle are the same thing, energy is conserved, and the gain in Kinetic Energy is a result of loss of Potential Energy which is the same thing as the loss of Rest mass.
The implication, if correct, is that the Rest mass Energy of a particle falling from infinity is completely converted to Kinetic Energy by the time it gets to the event horizon. In effect the falling massive particle becomes a photon at the event horizon which is consistent with the particle attaining the speed of light by the time it gets to the event horizon. If it is true that a particle has no rest mass as it reaches the event horizon then that would be something very difficult to transform away by use of some alternative metric. For example it would not be possible to have a physical observer (with rest mass) fall smoothly and continuously through the event horizon. If it is not true, the equation for PE in the Woodhouse textbook is wrong OR Baez's claim that energy is conserved in Schwarzschild coordinates is wrong, OR the equation for the terminal velocity of a falling particle from the mathpages website is wrong OR my calculations are wrong.
It should be noted that the Potential Energy of a particle goes to zero at the event horizon and the final velocity goes to c at the event horizon for ANY initial height that the particle is dropped from, not just infinity.
Also, as mentioned in previous thread, a photon has no Rest mass Energy and as such it would seem that it is not possible for a photon to have gravitational Potential Energy, so if energy is conserved in Schwarzschild coordinates the Total Energy of a photon does not change as it rises or falls.
Those are my observations on the subject. I look forward to the members of this forum telling me where I have gone wrong.
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This question about potential energy seems similar to what I discussed in https://www.physicsforums.com/showthread.php?t=311563"
Gravitational blueshift corresponds to the difference between the underlying clock rates at the emitter and at the observer. Which means that the wavelength at the instant of emission is already longer as measured by the observer's clock than the same emission as measured by the emitter's clock.
On the other hand, without reference to clock rate differentials, as Einstein first pointed out, energy conservation requires the light beam to gain (kinetic) energy as it "falls" deeper into a gravity well. This causes the wavelength to decrease, so an observer located deeper inside the gravity well measures blueshift.
Obviously these two effects are complementary and not cumulative: the resulting gravitational blueshift is not twice the amount of the gravitational time dilation. They are flip sides of a single coin.
... is exactly right.
To claim that clocks lower down run slower than clocks higher up, is the same as claiming the frequency of a falling photon does not change...
... and to claim that the frequency of a photon increases as it falls, is the same as saying clocks higher up and lower down run at the same rate.
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If it is true that a particle has no rest mass as it reaches the event horizon then that would be something very difficult to transform away by use of some alternative metric.
That's just another coordinate singularity. It's measuring kinetic energy with lightspeed observers (thus very large) and then rescaling to a finite value.
Always bear in mind that at the EH, many Schwarzschild quantities are defined via an impossible set of observers. Simply expect impossible observers to measure impossible things.
That has nothing to do with the properties of the massive falling body.
I'd like to mention that $$E = m_0 c^2 \sqrt{1-2GM/rc^2}/\sqrt{1-v^2/c^2}$$ is conserved. Anything else is the work of the evil one
The Total Energy (TE) of a particle is given by:
$$TE = \gamma (RE) = \sqrt{(ME)^2+(RE)^2}$$
where Rest mass Energy $(RE) = m_o c^2$, Momentum Energy (ME) = pc and momentum $(p) = \gamma m_o v$.
Does this hold in Schwarzschild space? I was thinking something like TE=KE+PE, like Eq 7.47 of http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html.
The Total Energy (TE) of a particle is given by:
$$TE = \gamma (RE) = \sqrt{(ME)^2+(RE)^2}$$
where Rest mass Energy $(RE) = m_o c^2$, Momentum Energy (ME) = pc and momentum $(p) = \gamma m_o v$.
Does this hold in Schwarzschild space? I was thinking something like TE=KE+PE, like Eq 7.47 of http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll7.html.
I should have made it clear that I had substituted $m_o*\sqrt{1-2GM/(rc^2)}$ for $m_o$ in the original $TE = \gamma (RE) = \sqrt{(ME)^2+(RE)^2}$ equation when I proposed that PE is substituted for RE to make the energy balance in Schwarzschild coordinates.
The link you posted is very helpful but a little difficult to follow as it seems to assume the reader is already an expert in General Relativity. That reference defines the effective potential as:
$$\frac{E^2}{2} = \frac{1}{2} \left(\frac{\delta r}{\delta \lambda} \right)^2 + V(r)$$
By substituting equation 7.48 given in your link for V(r) and removing the terms that contain L as I am not concerned with angular momentum here, the effective potential is:
$$\frac{E^2}{2} = \frac{1}{2} \left(\frac{\delta r}{\delta \lambda} \right)^2 + \frac{1}{2}\epsilon-\epsilon \frac{GM}{r}$$
To obtain an equation that is in the correct units for energy, I have reintroduced the not explicitly stated $m_o$ and $c^2$ variables and then multiplied by 2 and taken the square root on both sides to obtain an expression for total energy :
$$TE = m_o c^2 * \sqrt{\frac{1}{c^2} \left(\frac{\delta r}{\delta \lambda} \right)^2 + \epsilon-\epsilon \frac{2GM}{r c^2}$$
Earlier in the link given by atyy just after equation 7.39 it states that for a massive particle we should use $\lambda = \tau$ (proper time of the falling particle) and $\epsilon = 1$ so I will use those parameters to give:
$$TE = m_o c^2 * \sqrt{\frac{1}{c^2} \left(\frac{\delta r}{\delta \tau} \right)^2 + \left(1-\frac{2GM}{r c^2} \right)$$ (Eq 1a)
where ${\delta r}/{\delta \tau}$ is the instantaneous radial falling velocity measured in proper time and $(1-2GM/(r c^2))$ is the familiar gravitational gamma factor squared and inverted.
The equation for ${\delta r}/{\delta \tau}$ is conveniently given by mathpages here http://www.mathpages.com/rr/s6-07/6-07.htm as
$$\frac{\delta r}{\delta \tau} = - \sqrt{\frac{2GM}{r} - \frac{2GM}{R}}$$
where R is the release height of the particle and r is the height the particle has fallen to.
By substituting this new expression for the radial proper falling velocity into (Eq 1a) the equation for Total Energy becomes:
$$TE = m_o c^2 * \sqrt{ \left( \frac{2GM}{r c^2} - \frac{2GM}{R c^2} \right) + \left(1-\frac{2GM}{r c^2}\right)$$ (Eq 1b)
For any initial release height (R) the TE given by that equation is constant for any r. This is especially obvious when R is set to infinity so that term goes to zero and the equation simplifies to:
$$TE = m_o c^2 * \sqrt{1}$$
If we use the symbol 'u' to represent the instantaneous radial falling velocity measured in proper time ${\delta r}/{\delta \tau} = - \sqrt{\left({2GM}/{r} - {2GM}/{R}\right)}$ then the Carroll TE equation (Eq 1a) can be expressed as :
$$TE = m_o c^2 * \sqrt{\frac{u^2}{c^2} + \left( 1-\frac{2GM}{r c^2} \right)$$ (Eq 1c)
It turns out that the equation given by Carroll (Equation 1a, 1b or 1c) is numerically equal to the equation by quZx which I will call (Eq 2a) :
I'd like to mention that $$E = m_0 c^2 \sqrt{1-2GM/rc^2}/\sqrt{1-v^2/c^2}$$ is conserved. Anything else is the work of the evil one
,when v is calculated using this mathpages equation given in my earlier post:
$$v = \sqrt{ 1 - \left(1-\frac{2GM}{r c^2}\right) \left(1-\frac{2GM}{R c^2}\right)^{-1}}$$
Returning to the original equation I gave in OP of this thread:
$$TE = \sqrt{(ME)^2+(RE)^2}$$
where RE is Rest Energy and ME is Momentum Energy and using the assumption that Potential Energy (PE) is synonomous with Rest Energy in the context of a particle in a gravitational field, then:
$$TE = \sqrt{(ME)^2+(PE)^2}$$ (Eq 3a)
which can be expanded to the more explicit equation:
$$TE = \sqrt{\frac{(1-2GM/(r c^2)) m_o^2 v^2 c^2}{(1-v^2/c^2)} +(1-2GM/(r c^2)) m_o^2 c^4}$$
Using the symbols $\gamma$ for the velocity gamma factor and g for the gravitational gamma factor the equation can be stated as:
$$TE = \sqrt{\left(\frac{\gamma m_o v c}{g} \right)^2 +\left (\frac{m_o c^2}{g}\right)^2}$$ (Eq 3b)
Similarly, the equation given by quZz can be expressed as:
$$TE = \frac{\gamma m_o c^2}{g}$$ (Eq 2b)
The equations 1a, 1b, 1c, 2a, 2b, 3a and 3b are all numerically equal and all agree that the total energy of a falling particle is constant and Kinetic Energy is gained at the expense of Rest Energy/Potential energy of the particle. At the event horizon the Rest Energy/Potential Energy is zero and all the energy of the particle is purely Kinetic.
It is also correct that:
TE = KE + PE as pointed out by atyy which is just another way of expressing the more familiar:
TE = KE + RE of special relativity.
The Caroll equations (1a, 1b and 1c) are superior only in that they do not suffer from division by zero errors for values of r <= 2GM/c^2 but the other equations may be more convenient in certain situations.
It is also worth noting that effective potential is not the same thing as potential energy and the two are often confused.
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I should have made it clear that I had substituted $m_o*\sqrt{1-2GM/(rc^2)}$ for $m_o$ in the original $TE = \gamma (RE) = \sqrt{(ME)^2+(RE)^2}$ equation when I proposed that PE is substituted for RE to make the energy balance in Schwarzschild coordinates.
Nice algebra! I'm pretty sure from other considerations that a massive particle does not turn into a photon at the event horizon, but am not sure what a correct way to answer your question is. How about considering that what one calls RE, PE etc is arbitrary - I suspect that only TE has a coordinate independent meaning as a conserved quantity related to a Killing vector. So how about trying some renaming (using ?) to think of sqrt(1-2GM/r) as representing RE?+PE?, so that RE? is constant, and PE? varies with r, and it just so happens that RE?+PE? is zero at the event horizon, but RE? is still non-zero, and so there is no massive particle turning into a photon?
Nice algebra! I'm pretty sure from other considerations that a massive particle does not turn into a photon at the event horizon, but am not sure what a correct way to answer your question is. How about considering that what one calls RE, PE etc is arbitrary - I suspect that only TE has a coordinate independent meaning as a conserved quantity related to a Killing vector. So how about trying some renaming (using ?) to think of sqrt(1-2GM/r) as representing RE?+PE?, so that RE? is constant, and PE? varies with r, and it just so happens that RE?+PE? is zero at the event horizon, but RE? is still non-zero, and so there is no massive particle turning into a photon?
I think I understand what you are getting at and I think you are on the right track. It occurred to me that even though the RE and/or PE of a massive particle goes to zero at the EH, the Rest Mass $m_o$ of a massive particle is always non-zero even at the event horizon, while the Rest mass of a photon is always exactly zero, so perhaps it would be unsafe to consider that a massive particle becomes a photon at the EH. It also occurs to me that the Potential Energy is by definition the stored energy that has potential to be converted to kinetic energy by nuclear processes or by virtue of the particles location in the gravitational field. When a particle arrives at the EH its coordinate velocity goes to zero as does the PE. This suggests there are no potential energy reserves for the particle to continue anywhere unless we are talking about going into negative potential energy and where is this energy borrowed from?
There are still some paradoxical things about the EH. For example a particle with zero rest mass can never be stationary and yet a photon can be stationary (with respect to the centre of the gravitational field) at the EH. A particle with non-zero Rest Mass can never travel at the speed of light and yet it can at the EH but it just so happens that the speed of light is zero there. At the EH all the energy of of a massive particle is kinetic and yet it is not moving in coordinate terms.
Anyway, it seems that even when the PE or RE of a massive particle is zero at the EH, it still retains the property of non-zero Rest Mass and that the information of being a massive particle is retained and recovered when the particle is eventually released as Hawking radiation when the black hole evaporates... so no information loss... maybe...
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I'm referring to Taylor & Wheeler's textbook Exploring Black Holes, which specifically says that a Newtonian analysis (potential energy goes down as kinetic energy increases) cannot be used to accurately assess a particle's relativistic energy in freefall in a gravitational field. It says that a "shell" observer (located on a theoretical mechanical shell surrounding the BH outside the EH) can perform a locally valid measurement of a falling particle's relativistic energy. This measurement is quite different than the energy calculated for the same falling particle by a very distant observer, which is always equal to the free falling particle's rest mass regardless of velocity. As you know, it is not valid for a distant observer to separate a particle's total energy into potential energy and kinetic energy components, nor into rest energy and kinetic energy components. On the other hand, it is valid for a shell observer to divide the particle's total energy into kinetic energy and rest energy components (but not a potential energy component). The shell observer will locally measure the particle's kinetic energy, and therefore its total energy, to increase as the particle falls farther toward the EH. The difference between the total energy measured locally by the shell observer and globally by an observer at infinity is:
$$\frac{E_{shell}}{E_{infinity}} = \frac{1}{\sqrt{1 - \frac{2GM}{c^{2}r}}}$$
Every shell observer, no matter how close to the EH, will locally measure light to be traveling at exactly the speed of light; not coming slowly to a rest at the EH. The shell observer will measure an infalling massive particle to approach the speed of light as it approaches the EH, not slow down.
Maybe I'm misreading your posts, but it seems to me you tend to shift your reference frame at various points by using distant observer attributes sometimes and shell observer attributes at other times, in the course of your calculations. If you are actually doing that (I'm not entirely sure), you will obtain seemingly paradoxical results. There is no doubt that relativistic analysis is abstract because the calculated and measured results are so different depending on the observer's situation.
I believe it also is fundamentally invalid to introduce the concept of potential energy into relativistic equations such as the Schwarzschild metric. At best, PE can be treated as a Newtonian analogy for only limited parts of the relativistic calculations.
The difference between the total energy measured locally by the shell observer and globally by an observer at infinity is:
$$\frac{E_{shell}}{E_{infinity}} = \frac{1}{\sqrt{1 - \frac{2GM}{c^{2}r}}}$$
Agree, but ratio going to infinity at the EH is a nuisance.
Every shell observer, no matter how close to the EH, will locally measure light to be traveling at exactly the speed of light; not coming slowly to a rest at the EH. The shell observer will measure an in-falling massive particle to approach the speed of light as it approaches the EH, not slow down.
This is the equation I gave for the velocity used in the equations:
$$v = \sqrt{ 1 - \left(1-\frac{2GM}{r c^2}\right) \left(1-\frac{2GM}{R c^2}\right)^{-1}}$$
You will notice that if r is set to 2GM/c^2 that v=1 and so this equation is the local measurement of velocity. I modified the original coordinate velocity of a falling particle given in mathpages http://www.mathpages.com/rr/s6-07/6-07.htm as:
$$v = \left(1-\frac{2GM}{r c^2} \right) \sqrt{ 1 - \left(1-\frac{2GM}{r c^2}\right) \left(1-\frac{2GM}{R c^2}\right)^{-1}}$$
by removing the $\left(1-{2GM}/{r c^2} \right)$ factor on the left of the square root term precisely for the purpose of expressing velocity in local terms rather than coordinate terms.
Maybe I'm misreading your posts, but it seems to me you tend to shift your reference frame at various points by using distant observer attributes sometimes and shell observer attributes at other times, in the course of your calculations.
I am aware of that possibility and I am exploring it further, but if I am guilty of doing that, then so is Sean M. Carroll here http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll_contents.html and N.M.J. Woodhouse on page 100 of his textbook "General Relativity". Woodhouse gives the Total Energy of a falling particle as :
$$E = \frac{\sqrt{1-2M/r}}{\sqrt{1-v^2}}$$
which is essentially the same as the equation given by quZz in his earlier post.
I agree that a local observer is unable to determine the gravitational gamma factor without reference to the measurements of other observers not located at his locality. I also agree that if the coordinate velocity is used then it is difficult to make an energy balance that conserves energy because the coordinate velocity goes to zero as the EH is approached.
I also admit I am not sure of the exact nature of potential energy in GR and that is the reason for starting this thread, so that I might learn more.
I would also be interested if anyone could post an equation and or derivation ( a link will do) for gravitational potential in GR as I can not seem to find much information on that subject.
[EDIT] It occurred to me after posting this, that the falling velocity could be expressed in coordinate terms as long as the speed of light is also expressed in coordinate terms, i.e. $c_{(COORD)} = c_{(LOCAL)}(1-2GM/(r c^2))$. That would probably work out fine and the Total Energy expression would then be completely in coordinate terms rather than the mixed coordinate and local terms, which is a valid concern.
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... On the other hand, it is valid for a shell observer to divide the particle's total energy into kinetic energy and rest energy components (but not a potential energy component).
Sure, a local observer can measure the local falling velocity, but I am curious how he would determine the rest mass of the falling object? A mass of 2kgs at infinity and a mass of 1kg at infinity fall at exactly the same rate so the local falling velocity tells you nothing about the rest mass of the falling object.
Maybe the falling objects can be arranged to collide with lightly suspended stationary test masses at a given radius and the mass of the falling object estimated from momentum considerations?
... The shell observer will locally measure the particle's kinetic energy, and therefore its total energy, to increase as the particle falls farther toward the EH.
To be sure that the total energy is increasing, the local observer would have to have complete knowledge of the all the energy components of the particle.
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Please be more specific about what you think would be wrong in Carroll or Woodhouse. The link to Carroll isn't specific.
Is there a place where they mix the perspectives of an observer at infinity (usual for the Scwharzschild metric) and a shell observer?
Please be more specific about what you think would be wrong in Carroll or Woodhouse. The link to Carroll isn't specific.
Is there a place where they mix the perspectives of an observer at infinity (usual for the Scwharzschild metric) and a shell observer?
No, I am not saying that there is anything wrong with the Carroll or Woodhouse equations. I am saying their equations numerically agree with mine, so if mine is wrong, so are theirs. However, you are right that I was guilty of introducing a local measurement of velocity, but that did not affect the results, as I was using a local measurement for the speed of light also.
$$\frac{v^2_{(COORD)}}{c^2_{(COORD)}} = \frac{ v^2_{(LOCAL)}(1-2M/r)^2} { c^2_{(LOCAL)}(1-2M/r)^2} = \frac{v^2_{(LOCAL)}}{c^2_{(LOCAL)}}$$
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The analysis of short-run production is commonly performed at the introductory level with simple tables and graphs. These are useful abstraction methods for isolating and analyzing key aspects of short-run production. However, mathematical equations are another, often more powerful, method of abstract analysis. This is where the production function comes into play. Because a mathematical equation can extend beyond the two dimensions of a graph, it is possible to consider relations beyond just that for a variable input and total production.
If the production function takes the form of a specific equation (such as Q = 5L + 10K + 2LK), then a total product curve relating total product and the variable input can be plotted. However, to do so, one of the two inputs (L or K) must be designated as a variable input and one designated as a fixed input. For most types of production, labor is more readily changed than capital, so L is generally the variable input and K is usually the fixed input. A short-run total product curve can then be derived by "fixing" K at a particular value, then plotting the values of Q for alternative values of L.
While a great deal of economic insight into short-run production decisions of a firm and market supply curves can be analyzed with a simple graph, when economists begin using mathematical equations, such as the production function, Q = f(L, K), the possibilities are almost unlimited. A wide assortment of additional input variables can be added to this equation to make it, not only more sophisticated, but also more revealing.
For example, the effect of education and human capital on production can be seen by adding the educational attainment of workers as another input variable. In addition, the alternative impact on production of different types of capital, such as fixed structures and equipment can be identified by separating capital into two variables. In addition, to identify how public infrastructure, like highways and streets, affects production, then a variable for this input can be added to the production function.
<= PRODUCTION COST PRODUCTION INPUTS =>
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Thanks for visiting AmosWEB | 884 | 4,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-30 | latest | en | 0.925522 |
https://cstheory.stackexchange.com/questions/5269/sdp-relaxation-of-independent-set?noredirect=1 | 1,601,436,756,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402101163.62/warc/CC-MAIN-20200930013009-20200930043009-00528.warc.gz | 309,587,828 | 36,492 | # SDP relaxation of independent set
I'm looking at page 28 of Lovasz "Semidefinite programs and combinatorial optimization" and it gives the following approximation of independence number of the graph
$$\max u' Z u$$ subject to $$Z\succ 0$$ $$Z_{ij}=0 \ \forall ij\in E(G)$$ $$tr(Z)=1$$
Can I get independent set (or something close to an independent set) directly from the solution of SDP relaxation? Lovasz says that SDP is the only known way to solve this problem exactly for perfect graphs, is that still true?
Clarification: there's a similar SDP relaxation for the size of maximum cut, and I can get the full solution (the actual cut, rather than its size) by taking square root of Z and doing randomized rounding (Ch.6 of Williamson/Shmoys book). I'm wondering if there's a similar technique for this problem
• For the first question, I don't really get what you mean by "the actual independent set". The SDP is a relaxation, and so the optimal value of the SDP bounds the independence number from above. If they differ, no independent set attains the optimal value of the SDP. This can be the case if the graph is not perfect. Could you make it more explicit what you require for your "actual independent set"? – Yoshio Okamoto Mar 4 '11 at 0:25
• I want to get largest independent set rather than "size of largest independent set" – Yaroslav Bulatov Mar 4 '11 at 1:15
• Thanks for clarification, but I'm still wondering. The SDP for max cut is used for approximation. Namely, the randomized rounding gives a cut that has value "close" to the optimal cut value, not necessarily a real max cut. If you need a similar technique, I guess what you really want is an independent set that has size close to the independence number. Or, do you concentrate on perfect graphs, or want to deal with general graphs? – Yoshio Okamoto Mar 4 '11 at 1:38
• I want to find Maximum Independent Set in Perfect Graph. ipsofacto gives a solution, but it requires solving several SDPs – Yaroslav Bulatov Mar 4 '11 at 2:11
I believe SDP is the only known technique to solve the maximum independent set problem on perfect graphs. To get the independent set, one could do the following. Guess if a vertex is in the independent set, delete it and solve the SDP. If it returns the same value, then there is an independent set without this vertex. So, make this vertex adjacent to all other vertices, and continue. This should give you an actual independent set.
Otherwise, we have identified one vertex of the independent set, and we can remove it and continue on the remaining graph.
• Moreover, this has been implemented and works quite well (with some optimizations): E. Alper Yıldırım and Xiaofei Fan-Orzechowski, On Extracting Maximum Stable Sets in Perfect Graphs Using Lovász's Theta Function, Computational Optimization and Applications 33, 229–247, 2006. dx.doi.org/10.1007/s10589-005-3060-5 – András Salamon Mar 4 '11 at 11:03
• Interesting...it seems that perfection is not required for SDP estimate of independence number to be exact (here's example mathoverflow.net/questions/57336/…), so this should work for a larger class of graphs – Yaroslav Bulatov Mar 4 '11 at 20:33
• @Yaroslav: You're right, perfection is not required. But if you adapt the strategy ipsofacto suggested, you will need the deletion of vertices also has the same property. This condition is automatically satisfied if the graph is perfect, but otherwise you need to be careful. – Yoshio Okamoto Mar 5 '11 at 10:40
I'm not sure if Lovasz's comment still holds. There has been some recent work on this (and related) problems on perfect graphs. You should take a look at the following link for techniques that involve message passing rather than solving SDPs: http://www.cs.columbia.edu/~jebara/papers/uai09perfect.pdf
• Interesting paper, do I understand it correctly that if max-product converges on a perfect graph, then greedy decoding will recover maximum independent set? – Yaroslav Bulatov Mar 4 '11 at 4:12
• I've skimmed the paper, but I couldn't find how the methods solve the maximum independent set problem for perfect graphs in polynomial time. The number of maximal cliques can be exponential in a perfect graph, and so running times in Corollaries 1 and 2 are not polynomial. Although I don't understand the contents of Section 7 very much, I don't see which linear optimization problem the method in Section 7 solves. The experiments are performed for the maximum matching problem, but not for the maximum independent set problem. – Yoshio Okamoto Mar 4 '11 at 4:43
• @yoshio You are correct. The LP for MWIS is known to be integral if you include the appropriate constraints over all of the (exponentially many) cliques. And it is the perfectness of the clique graph that the paper discusses. It looks like the authors only conjecture that max-product on a NMRF always produces the correct MAP assignment. – Nicholas Ruozzi Mar 4 '11 at 14:51
• Thanks. Then, can I assume that the paper does not give a polynomial-time algorithm for the maximum independent set problem for perfect graphs? – Yoshio Okamoto Mar 5 '11 at 10:42
• @YoshioOkamoto: it seems so. A recent paper gives an example of a perfect graph where this approach converges to wrong solution. Figure 3 of "Revisiting MAP Estimation, Message Passing and Perfect Graphs" (datalab.uci.edu/papers/AISTATS_perfect_graphs.pdf) – Yaroslav Bulatov Sep 20 '11 at 0:59 | 1,326 | 5,413 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-40 | latest | en | 0.9219 |
https://socratic.org/questions/how-do-you-write-the-standard-form-of-the-equation-through-3-1-perpendicular-to-#332069 | 1,721,532,612,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517550.76/warc/CC-MAIN-20240721030106-20240721060106-00313.warc.gz | 452,321,647 | 6,618 | # How do you write the standard form of the equation through: (3,1), perpendicular to y=-2/3x+4?
Nov 6, 2016
$3 x - 2 y = 7$
#### Explanation:
Write the standard form of the line that goes through $\left(3 , 1\right)$ and is perpendicular to $y = - \frac{2}{3} x + 4$.
The equation $y = \textcolor{red}{- \frac{2}{3}} x + 4$ is in slope intercept form
$y = \textcolor{red}{m} x + b$ where $\textcolor{red}{m}$= slope and $b$ = the $y$ intercept.
The slope of this line is then $m = \textcolor{red}{- \frac{2}{3}}$
A perpendicular slope is the opposite sign reciprocal. So, we change the sign of $\textcolor{red}{- \frac{2}{3}}$ and switch the numerator and denominator.
Perpendicular slope $\textcolor{b l u e}{m} = \textcolor{b l u e}{\frac{3}{2}}$
To find the equation of the new line, use the point slope equation
$y - {y}_{1} = m \left(x - {x}_{1}\right)$ where $m =$ slope and $\left({x}_{1} , {y}_{1}\right)$ is a point.
The slope is $\textcolor{b l u e}{\frac{3}{2}}$ and the point is the given point $\left(3 , 1\right)$.
$y - 1 = \textcolor{b l u e}{\frac{3}{2}} \left(x - 3\right) \textcolor{w h i t e}{a a a}$Distribute
$y - 1 = \frac{3}{2} x - \frac{9}{2}$
Standard form is $a x + b y = c$ where $a , b \mathmr{and} c$ are integers and $a$ is positive.
$\textcolor{w h i t e}{a a} 2 \left(y - 1 = \frac{3}{2} x - \frac{9}{2}\right) \textcolor{w h i t e}{a a a}$Multiply the equation by $2$
$\textcolor{w h i t e}{a a a a a} 2 y - 2 = 3 x - 9$
$- 3 x \textcolor{w h i t e}{a a a a a a a} - 3 x \textcolor{w h i t e}{a a a}$Subtract $3 x$ from both sides
$- 3 x + 2 y - 2 = - 9$
$\textcolor{w h i t e}{a a a a a a a a} + 2 \textcolor{w h i t e}{a a a} + 2 \textcolor{w h i t e}{a a a}$Add $2$ to both sides
$- 3 x + 2 y = - 7$
$- 1 \left(- 3 x + 2 y = - 7\right) \textcolor{w h i t e}{a a a}$Multiply the equation by $- 1$
$3 x - 2 y = 7$ | 748 | 1,867 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 33, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.96875 | 5 | CC-MAIN-2024-30 | latest | en | 0.614835 |
https://meeb.us/what-is-a-variable-in-math-definition/ | 1,618,859,007,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038916163.70/warc/CC-MAIN-20210419173508-20210419203508-00604.warc.gz | 479,382,953 | 10,722 | What Is A Variable In Math Definition
They are useful for finding. In algebra a variable is an alphabet which is used to represent the unknown number.
Coefficient Variable Constant Equation Expression Definition Google Search Math Pictures Expression Definiti Math Pictures Expression Definition Exponents
In x 2 6 x is the.
What is a variable in math definition. Variable definition in maths. Making algebraic computations with variables as if they were explicit numbers allows one to solve a range of problems in a single computation. M for the amount of mass used and 3.
In addition to numbers variables are commonly used to represent vectors matrices and functions. In mathematics a variable is a quantity that can change. Splashlearn is an award winning math learning program used by more than 30 million kids for fun math practice.
Variables stand for things that you want to find but don t have the answer to yet. E for the amount of energy produced 2. A typical example is the quadratic formula which allows one to solve every quadratic e.
In other words a variable is a symbol for a number where the value is not known. Einstein s famous equation e mc 2 uses the following variables. In mathematics a variable is a symbol which functions as a placeholder for varying expression or quantities and is often used to represent an arbitrary element of a set.
Letters are used to represent these changing unknown quantities. C 2 to represent the speed of light squared. It represents the value.
Also learn the facts to easily understand math glossary with fun math worksheet online at splashlearn. Really variables are used wherever a best answer needs to be found. The generic letters which are used in many algebraic expressions and equations are x y z.
A variable is a quantity that may be changed according to the mathematical problem. Definition of variable explained with real life illustrated examples. A symbol for a value we dont know yet.
It is usually a letter like x or y. Illustrated definition of variable.
In This Page We Are Going To Discuss About Independent Variable Below You Can See How We Define Inde Dependent And Independent Variables Variables Independent
Monomial Definition Math Definitions Expressions
In An Algebraic Equation Independent Variable Means A Variable Whose Values Are Independent Of Dependent And Independent Variables Variables Algebra Equations
Definition Of Variable And Examples Math Examples Variables Definitions
The Algebraic Expressions Have The Variables And The Constants The Algebraic Expressions Are The F Algebraic Expressions Math Strategies Expression Definition
Image From Https Www Mathsisfun Com Algebra Images Variable Constant Gif Algebra Equations Algebra High School Help | 512 | 2,754 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2021-17 | latest | en | 0.912699 |
https://brainmass.com/math/basic-algebra/4590 | 1,685,318,942,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644571.22/warc/CC-MAIN-20230528214404-20230529004404-00739.warc.gz | 179,426,599 | 75,165 | Explore BrainMass
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This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Solve the equation x^2 + 3x - 10 = 0 by factoring.
https://brainmass.com/math/basic-algebra/4590
#### Solution Preview
Original Equation
x^2 + 3x - 10 = 0
Factor
(x + 5)(x - 2) = 0
Set each factor equal to 0
x + 5 = 0 or x - 2 ...
#### Solution Summary
A quadratic equation is solved by factoring.
\$2.49 | 160 | 546 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.09375 | 3 | CC-MAIN-2023-23 | longest | en | 0.873527 |
https://discussions.unity.com/t/calculate-a-new-vector-which-is-20-degrees-towards-a-new-vector-from-an-old-vector/51695 | 1,708,643,662,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473871.23/warc/CC-MAIN-20240222225655-20240223015655-00826.warc.gz | 217,144,206 | 5,759 | # Calculate a new vector which is 20 degrees towards a new vector from an old vector
hi there having a problem calculating a new vector mainly due to my failing of understanding of vectors and Quaternions.
i have 3 points A, B and C and two vectors A->B and B->C,
i want to create a new vector which is vector AB rotated by 20 degrees towards vector BC.
what i have tried to do is get the cross product between the two vector to get a rotation axis then use Quaterion.AngleAxis, however i don’t know how to apply this to a vector3.
how would i go about doing this correctly?
thanks for any help
Ken.
here is my bad code if it helps
``````Vector3 newWirePoint;
Vector3 A = wirePoint[x-1]-wirePoint[x-2];
Vector3 B = anchorPos-wirePoint[x-1];
Vector3 AcrossB = Vector3.Cross(A,B);
newWirePoint = A.normalized * 0.05f;
newWirePoint = Quaternion.AngleAxis(20,AcrossB);
``````
You are nearly there!
`````` newWirePoint = Quaternion.AngleAxis(20, AcrossB) * newWirePoint;
``````
Found my own answer will leave post here so if anyone else has this issue.
here is my script
`````` float Omega = 0.35f;
Vector3 newWirePoint;
Vector3 A = wirePoint[x-1]-wirePoint[x-2];
Vector3 B = anchorPos-wirePoint[x-1];
Vector3 AcrossB = Vector3.Cross(A,B);
AcrossB = AcrossB.normalized;
newWirePoint = A*Mathf.Cos(Omega)+ Vector3.Cross(AcrossB,A)*Mathf.Sin(Omega)+ AcrossB*Vector3.Dot(AcrossB,A)*(1-Mathf.Cos(Omega));
newWirePoint = newWirePoint.normalized * 0.05f;
wirePoint[x] = wirePoint[x-1] + newWirePoint;
wireRender.SetVertexCount(x+1);
wireRender.SetPosition(x,wirePoint[x-1] + newWirePoint);
x++; | 449 | 1,600 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.46875 | 3 | CC-MAIN-2024-10 | latest | en | 0.814015 |
https://www.philosophy-science-humanities-controversies.com/listview-details.php?id=260364&a=a&first_name=Bertrand&author=Russell&concept=Space | 1,643,345,667,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305420.54/warc/CC-MAIN-20220128043801-20220128073801-00589.warc.gz | 990,809,641 | 5,038 | Philosophy Dictionary of Arguments
Home
Space, philosophy: various discussions deal, among others, with the question whether the space is absolute or whether empty space is possible. In different sciences, multi-dimensional spaces with certain properties are used to better calculate like Hilbert spaces in the theory of relativity or multidimensional spaces in mathematical nodal theory. No ontological assumptions are made. See also substantivalism, relativism, movement, absoluteness, compactness, conceptual space, dimensions, logical space, four-dimensionalism.
_____________
Annotation: The above characterizations of concepts are neither definitions nor exhausting presentations of problems related to them. Instead, they are intended to give a short introduction to the contributions below. – Lexicon of Arguments.
Author Concept Summary/Quotes Sources
Bertrand Russell on Space - Dictionary of Arguments
Kursbuch 8 Mathematik, Frankfurt/M. März 1967
12
Weierstraß/Russell: banished the infinitesimal variable from mathematics - VsZenon: error: to think that the world must always be the same because there is no state of change. - Without infinitesimal variables: time: there is no more "next moment". - No immediate result of two moments - in between there are ever more moments. - Space: ditto: is always further divisible.
B. Russell, ABC of Relativity Theory Frankfurt 1989
II 46
Def "space-like"/Russell: two events are space-like if it is impossible for a body to move fast enough to be present at both events - but it can be "halfway" and perceive both as happening at the same time.
Def "time-like"/Russell: two events are time-like if it is possible for a physical body to be present at both events. Borderline case: E.g. two events as part of a light beam or - E.g. an event. perception of the other event: then distance 0.
Def Distance/Russell: is a physical fact which is part of the events and does not depend on the circumstances of observer.
Neglecting gravitation, one can apply the Special Theory of Relativity. Then the distance between two events can be calculated if one knows the spatial and temporal distance, measured by an arbitrary observer.
_____________
Explanation of symbols: Roman numerals indicate the source, arabic numerals indicate the page number. The corresponding books are indicated on the right hand side. ((s)…): Comment by the sender of the contribution. Translations: Dictionary of Arguments
The note [Concept/Author], [Author1]Vs[Author2] or [Author]Vs[term] resp. "problem:"/"solution:", "old:"/"new:" and "thesis:" is an addition from the Dictionary of Arguments. If a German edition is specified, the page numbers refer to this edition.
Russell I
Principia Mathematica Frankfurt 1986
Russell II
B. Russell
The ABC of Relativity, London 1958, 1969
German Edition:
Das ABC der Relativitätstheorie Frankfurt 1989
Russell IV
B. Russell
The Problems of Philosophy, Oxford 1912
German Edition:
Probleme der Philosophie Frankfurt 1967
Russell VI
B. Russell
"The Philosophy of Logical Atomism", in: B. Russell, Logic and KNowledge, ed. R. Ch. Marsh, London 1956, pp. 200-202
German Edition:
Die Philosophie des logischen Atomismus
In
Eigennamen, U. Wolf (Hg), Frankfurt 1993
Russell VII
B. Russell
On the Nature of Truth and Falsehood, in: B. Russell, The Problems of Philosophy, Oxford 1912 - Dt. "Wahrheit und Falschheit"
In
Wahrheitstheorien, G. Skirbekk (Hg), Frankfurt 1996
> Counter arguments against Russell
Ed. Martin Schulz, access date 2022-01-28 | 839 | 3,518 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2022-05 | latest | en | 0.859414 |
https://oeis.org/A141692 | 1,623,754,459,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487620971.25/warc/CC-MAIN-20210615084235-20210615114235-00345.warc.gz | 393,052,360 | 4,970 | The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
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A141692 Triangle read by rows: T(n,k) = n*(binomial(n - 1, k - 1) - binomial(n - 1, k)), 0 <= k <= n. 3
0, -1, 1, -2, 0, 2, -3, -3, 3, 3, -4, -8, 0, 8, 4, -5, -15, -10, 10, 15, 5, -6, -24, -30, 0, 30, 24, 6, -7, -35, -63, -35, 35, 63, 35, 7, -8, -48, -112, -112, 0, 112, 112, 48, 8, -9, -63, -180, -252, -126, 126, 252, 180, 63, 9, -10, -80, -270, -480, -420, 0, 420, 480, 270, 80, 10 (list; table; graph; refs; listen; history; text; internal format)
OFFSET 0,4 COMMENTS The row sums are zero. Row n consists of the coefficients in the expansion of n*(x - 1)*(x + 1)^(n - 1). - Franck Maminirina Ramaharo, Oct 02 2018 LINKS G. C. Greubel, Table of n, a(n) for n = 0..5150 (Rows n=1..100 of triangle, flattened; offset adapted by Georg Fischer, Jan 31 2019) Rida T. Farouki, The Bernstein polynomial basis: A centennial retrospective, Computer Aided Geometric Design Vol. 29 (2012), 379-419. Eric Weisstein's World of Mathematics, Bernstein Polynomial Wikipedia, Bernstein polynomial FORMULA T(n,k) = n*(B(1/2;n-1,k-1) - B(1/2;n-1,k))*2^(n - 1), where B(t;n,k) = binomial(n,k)*t^k*(1 - t)^(n - k) denotes the k-th Benstein basis polynomial of degree n. T(n,k) = n*A112467(n,k). From Franck Maminirina Ramaharo, Oct 02 2018: (Start) T(n,k) = -T(n,n-k) T(n,0) = -n. T(n,1) = -A067998(n) E.g.f.: (x*y - y)/(x*y + y - 1)^2. Sum_{k=0..n} abs(T(n,k)) = 2*A100071(n). Sum_{k=0..n} T(n,k)^2 = 2*A037965(n). Sum_{k=0..n} k*T(n,k) = A001787(n). Sum_{k=0..n} k^2*T(n,k) = A014477(n-1). (End) EXAMPLE Triangle begins: 0; -1, 1; -2, 0, 2; -3, -3, 3, 3; -4, -8, 0, 8, 4; -5, -15, -10, 10, 15, 5; -6, -24, -30, 0, 30, 24, 6; -7, -35, -63, -35, 35, 63, 35, 7; -8, -48, -112, -112, 0, 112, 112, 48, 8; -9, -63, -180, -252, -126, 126, 252, 180, 63, 9; -10, -80, -270, -480, -420, 0, 420, 480, 270, 80, 10; ... MAPLE a:=proc(n, k) n*(binomial(n-1, k-1)-binomial(n-1, k)); end proc: seq(seq(a(n, k), k=0..n), n=0..10); # Muniru A Asiru, Oct 03 2018 MATHEMATICA Table[Table[n*(Binomial[n - 1, k - 1] - Binomial[n - 1, k]), {k, 0, n}], {n, 0, 12}]//Flatten PROG (Maxima) T(n, k) := n*(binomial(n - 1, k - 1) - binomial(n - 1, k))\$ tabl(nn) := for n:0 thru nn do print(makelist(T(n, k), k, 0, n))\$ /* Franck Maminirina Ramaharo, Oct 02 2018 */ CROSSREFS Cf. A007318, A112467, A128433, A128434. Sequence in context: A307297 A307301 A307300 * A261097 A335335 A261217 Adjacent sequences: A141689 A141690 A141691 * A141693 A141694 A141695 KEYWORD easy,tabl,sign AUTHOR Roger L. Bagula, Sep 09 2008 EXTENSIONS Edited, new name and offset corrected by Franck Maminirina Ramaharo, Oct 02 2018 STATUS approved
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Last modified June 15 04:47 EDT 2021. Contains 345043 sequences. (Running on oeis4.) | 1,378 | 3,191 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.34375 | 3 | CC-MAIN-2021-25 | latest | en | 0.377358 |
https://bugs.mojang.com/browse/MC-7200?attachmentViewMode=gallery | 1,709,052,780,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474676.79/warc/CC-MAIN-20240227153053-20240227183053-00711.warc.gz | 152,855,304 | 26,388 | ## Cave/tunnel generation may cut tunnels a bit too soon (fix included)
XMLWordPrintable
• Bug
• Resolution: Fixed
• Minecraft 1.4.6, Minecraft 1.4.7, Minecraft 1.5, Minecraft 1.6.1, Minecraft 1.6.2, Minecraft 1.7.5, Minecraft 14w11b, Minecraft 15w47c, Minecraft 1.10.2, Minecraft 16w42a, Minecraft 1.11, Minecraft 1.11.2, Minecraft 17w16b, 1.15.2, 20w07a, 20w14a, 20w17a, 20w21a, 1.16 Release Candidate 1, 1.16, 1.16.4, 21w08a, 21w08b, 21w13a, 21w14a, 21w16a
• World generation
• Normal
(Possible test cases moved under own heading below.)
Slightly related to MC-7196, same classes and methods in question, almost the same effect, but different bug in the code.
Cave/tunnel generator, both for Nether and overworld, may sometimes cut tunnels of at chunk borders, leaving flat straight walls, or have straight-angle incursions of stone into the tunnel. For this bug I do not have examples, because knowing which particular straight wall is because of this bug (rare) or because of the much more common MC-7196 is difficult.
However, explanation of mistake in the code might suffice...
Background
The tunnel generation algorithm has an optimization which calculates if the distance from the algorithm's current "tunnel digging" point to the generated chunk's center is longer than the remaining length of the tunnel. If the former length is longer, the tunnel can not reach the chunk, so the algorithm can be stopped there.
The bug
The distance between current digging point and chunk center is basically calculated correctly (though kept in the squared value), but the comparison is done with incorrect math, trying to subtract the square of the remaining length from the squared distance, and then compare the result to tunnel's maximum diameter (and a bit more) squared. Squared values can not be subtracted or compared like that.
If one does so (like in this case), the results will (incorrectly) vary depending on which particular relative locations are in question. Thus, at one chunk, the algorithm may think that the tunnel can not reach the chunk, while drawing the neighbour chunk, the algorithm thinks it will easily reach that chunk, and can correctly carve it, right the to edge of that chunk, leaving nasty straight cut at some chunk border.
Here is the bad code, with my variable name interpretations:
MapGenCavesHell.generateCaveNode(...)
```double xDeltaToChunkCenter = startX - chunkCenterX;
double zDeltaToChunkCenter = startZ - chunkCenterZ;
double remainingLength = (double)(fullLength - currentDistance);
double tunnelDiameterAndMore = (double)(diameterVariance + 2.0F + 16.0F);
if (xDeltaToChunkCenter * xDeltaToChunkCenter + ZDeltaToChunkCenter * ZDeltaToChunkCenter - remainingLength * remainingLength > tunnelDiameterAndMore * tunnelDiameterAndMore) {
return;
}
```
Exactly equal bug exists in the MapGenCaves class.
The fix
MapGenCavesHell.generateCaveNode(...)
```double xDeltaToChunkCenter = startX - chunkCenterX;
double zDeltaToChunkCenter = startZ - chunkCenterZ;
double remainingLength = (double)(fullLength - currentDistance);
double tunnelDiameterAndMore = (double)(diameterVariance + 2.0F + 16.0F);
double directDistanceSquared = xDeltaToChunkCenter * xDeltaToChunkCenter + zDeltaToChunkCenter * zDeltaToChunkCenter;
double tunnelsMaximumRemainingReachSquared = (remainingLength + tunnelDiameterAndMore) * (remainingLength + tunnelDiameterAndMore);
if (directDistanceSquared > tunnelsMaximumRemainingReachSquared) {
return;
}
```
That is, the math has been changed to compare the distance squared against remaining reach squared, which should be valid math.
Possible test cases
From Jens-Oliver Tofahrn, looks like possibly at least 3 "cuts" near each other; tp goes right in there, dark place, get torch:
Demo seed "North Carolina" or -343522682, F3-n, F3-g
/tp @s -48.42 18.94 0.10 -463.95 55.80
Multiple cuts near each other, the first place even has two cuts in the same view:
Seed 1479112774635546442
/tp @p -440 62 11 107 20
/tp @p -457 62 18 -140 19
/tp @p -447 62 -96 114 -12
(Wery Old: Semi-quick test case, though not 100% certain if it is caused by this issue: seed -4542366974610774625, Nether 218, -67, dig down to about 19 and look south east.)
History
This bug has been around for at least since summer 2011. Back then I found it, fixed it, reported it, and provided screenshots for "before and after". I estimated very roughly that in the overworld, about 1/4th of straight wall glitches in tunnels were caused by this bug, the rest by the other bug (MC-7196 when it was still effective for overworld, too), based on observed differences while enabling either fix separately.
1. cut1.png
506 kB
2. cut2.png
479 kB
Unassigned
Markku | 1,235 | 4,703 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-10 | latest | en | 0.873444 |
https://www.studiestoday.com/ncert-solution/ncert-class-12-physics-solutions-magnetism-and-matter-202481.html | 1,619,197,369,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039596883.98/warc/CC-MAIN-20210423161713-20210423191713-00627.warc.gz | 1,036,633,021 | 25,663 | # NCERT Class 12 Physics Solutions Magnetism And Matter
NCERT Class 12 Physics Solutions Magnetism And Matter- NCERT Solutions prepared for CBSE students by the best teachers in Delhi.
Class XII Chapter 5 – Magnetism And Matter Physics
Question 5.1: Answer the following questions regarding earth’s magnetism:
(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.
(b) The angle of dip at a location in southern India is about 18º. Would you expect a greater or smaller dip angle in Britain?
(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?
(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?
(e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 1022 J T−1 located at its centre. Check the order of magnitude of this number in some way.
(f ) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?
(a) The three independent quantities conventionally used for specifying earth’s magnetic field are:
(i) Magnetic declination,
(ii) Angle of dip, and
(iii) Horizontal component of earth’s magnetic field
(b)The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. The angle of dip would be greater in Britain (it is about 70°) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole.
(c)It is hypothetically considered that a huge bar magnet is dipped inside earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole. Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground.
(d)If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be free to move in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction.
(e)Magnetic moment, M = 8 × 1022 J T−1
Radius of earth, r = 6.4 × 106 m
Magnetic field strength, Where,
= Permeability of free space =
This quantity is of the order of magnitude of the observed field on earth.
(f)Yes, there are several local poles on earth’s surface oriented in different directions. A magnetised mineral deposit is an example of a local N-S pole.
Question 5.2: Answer the following questions:
(a) The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?
(b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?
(c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ‘battery’ (i.e., the source of energy) to sustain these currents?
(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past?
(e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?
(f ) Interstellar space has an extremely weak magnetic field of the order of 10−12 T. Can such a weak field be of any significant consequence? Explain.
[Note: Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.]
(a) Earthfs magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earthfs magnetic field with the time cannot be neglected.
(b)Earthfs core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not considered as a source of earthfs magnetism.
(c)Theradioactivity in earthfs interior is the source of energy that sustains the currents in the outer conducting regions of earthfs core. These charged currents are considered to be responsible for earthfs magnetism.
(d)Earth reversed the direction of its field several times during its history of 4 to 5 billion years. These magnetic fields got weakly recorded in rocks during their solidification. One can get clues about the geomagnetic history from the analysis of this rock magnetism.
(e)Earthfs field departs from its dipole shape substantially at large distances (greater than about 30,000 km) because of the presence of the ionosphere. In this region, earthfs field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them.
(f)An extremely weak magnetic field can bend charged particles moving in a circle. This may not be noticeable for a large radius path. With reference to the gigantic interstellar space, the deflection can affect the passage of charged particles.
Question 5.3: A short bar magnet placed with its axis at 30o with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 ~ 10.2 J. What is the magnitude of magnetic moment of the magnet?
Answer :Magnetic field strength, B = 0.25 T
Torque on the bar magnet, T = 4.5 ~ 10.2 J
Angle between the bar magnet and the external magnetic field,ƒÆ = 30‹
Question 5.4: A short bar magnet of magnetic moment m = 0.32 J T−1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Students should free download the NCERT solutions and get better marks in exams. Studiestoday.com panel of teachers recommend students to practice questions in NCERT books and download NCERT solutions.
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Current view: top level - Evolution/Systems/Ccz4 - RicciScalarPlusDivergenceZ4Constraint.hpp Hit Total Coverage Commit: 2ae2b99409ac582030d56a4560a92a3e066a7e54 Lines: 2 3 66.7 % Date: 2022-01-15 08:40:38 Legend: Lines: hit not hit
Line data Source code 1 0 : // Distributed under the MIT License. 2 : // See LICENSE.txt for details. 3 : 4 : #pragma once 5 : 6 : #include 7 : 8 : #include "DataStructures/Tensor/TypeAliases.hpp" 9 : #include "Utilities/Gsl.hpp" 10 : 11 : namespace Ccz4 { 12 : /// @{ 13 : /*! 14 : * \brief Computes the sum of the Ricci scalar and twice the divergence of the 15 : * upper spatial Z4 constraint 16 : * 17 : * \details Computes the expression as: 18 : * 19 : * \f{align} 20 : * R + 2 \nabla_k Z^k &= 21 : * \phi^2 \tilde{\gamma}^{ij} (R_{ij} + \nabla_i Z_j + \nabla_j Z_i) 22 : * \f} 23 : * 24 : * where \f$R\f$ is the spatial Ricci scalar, \f$Z^i\f$ is the upper spatial Z4 25 : * constraint defined by Ccz4::Tags::Z4ConstraintUp, \f$phi^2\f$ is the square 26 : * of the conformal factor defined by Ccz4::Tags::ConformalFactorSquared, 27 : * \f$\tilde{\gamma}^{ij}\f$ is the inverse conformal spatial metric defined by 28 : * Ccz4::Tags::InverseConformalMetric, \f$R_{ij}\f$ is the spatial Ricci 29 : * tensor defined by Ccz4::Tags::Ricci, and \f$\nabla_j Z_i\f$ is the gradient 30 : * of the spatial Z4 constraint defined by Ccz4::Tags::GradZ4Constraint. 31 : */ 32 : template 33 1 : void ricci_scalar_plus_divergence_z4_constraint( 34 : const gsl::not_null*> result, 35 : const Scalar& conformal_factor_squared, 36 : const tnsr::II& inverse_conformal_spatial_metric, 37 : const tnsr::ii& spatial_ricci_tensor, 38 : const tnsr::ij& grad_spatial_z4_constraint); 39 : 40 : template 41 1 : Scalar ricci_scalar_plus_divergence_z4_constraint( 42 : const Scalar& conformal_factor_squared, 43 : const tnsr::II& inverse_conformal_spatial_metric, 44 : const tnsr::ii& spatial_ricci_tensor, 45 : const tnsr::ij& grad_spatial_z4_constraint); 46 : /// @} 47 : } // namespace Ccz4
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http://www.tricki.org/article/Double_counting | 1,611,809,689,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610704835901.90/warc/CC-MAIN-20210128040619-20210128070619-00333.warc.gz | 182,342,668 | 21,158 | Tricki
## Double counting
### Quick description
If you have two ways of calculating the same quantity, then you may well end up with two different expressions. If these two expressions were not obviously equal in advance, then your calculations provide a proof that they are. Moreover, this proof is often elegant and conceptual. Double counting can also be used to prove inequalities, via a simple result about bipartite graphs.
### Example 1: sums of binomial coefficients
Prove that .
The left-hand side of this well-known identity counts all the subsets of size 0 from a set of size , then all the sets of size 1, then all the sets of size 2, and so on. By the time it has finished, it has counted all the subsets of a set of size . But an entirely different argument shows that the number of subsets of a set of size is : for each element there are two possibilities: belongs to the subset or does not belong to the subset. Therefore, in determining a subset of we have independent binary choices, so possibilities. (If this argument sounds a bit woolly, then you can prove it by induction. Let be the set . Then every subset of is obtained by taking a subset of and either adding to it or leaving it alone. Therefore, the number of subsets of is twice the number of subsets of . And the number of subsets of is 2.)
### Example 2: counting the edges of an icosahedron
Prove that the number of edges of an icosahedron is 30.
The idea of this example is to deduce the answer very simply from just the fact that an icosahedron has 20 triangular faces. The rough idea of the proof is that each face has three edges, while each edge is an edge of two faces, so the ratio of faces to edges ought to be 2 to 3. To be sure that this argument is correct, we count pairs , where is an edge and is a face that has as one of its edges. Since the icosahedron has 20 triangular faces, the number of such pairs is 60. If is the number of edges, then the number of such pairs is also , since each edge is part of two faces. Therefore, , as claimed.
### General discussion
Many combinatorial applications of the double-counting principle can be regarded as consequences of the following very simple result about bipartite graphs: if is a bipartite graph with vertex sets and , then the sum of the degrees of the vertices in is equal to the sum of the degrees of the vertices in . The proof is a very simple double count, since the two sums are different ways of counting the edges in (one by focusing on the vertex in and the other by focusing on the vertex in ).
For instance, for the icosahedron we could define a bipartite graph where one vertex set consisted of the edges of the icosahedron and the other of the faces, joining an edge to a face if the edge belonged to the face. The edges of this bipartite graph are precisely the edge-face pairs considered above (so this is not a different proof, but a different way of looking at the same proof).
The following equivalent reformulation of the bipartite-graphs principle can often be used very directly: if is the average degree of the vertices in and is the average degree of the vertices in , then .
Many applications of the principle use the following consequence. Let be a bipartite graph with vertex sets and . Suppose that every vertex in has degree and every vertex in has degree . Let be any subset of , and let be the set of all vertices in that are joined to at least one vertex in . Then . To prove this, let be the subgraph of with vertex sets and , with the edges inherited from . Then every vertex in has degree and every vertex in has degree at most .
It often happens that we are in the following situation. We have a set whose size we are trying to bound above, but we do not have an obvious bipartite graph to which to apply an inequality such as the one in the last paragraph. In that case, we must construct one. It is often easier to think about the task in an equivalent way: given a bipartite graph with vertex sets and , each vertex in determines a neighbourhood (defined to be the set of all vertices in that are joined to ), and therefore the bipartite graph itself determines a collection of subsets of , possibly with repeats, so strictly speaking it determines a multiset of subsets of . Conversely, given a collection of subsets of , we can use it to define a bipartite graph, by choosing for each set in the collection a vertex (or set of vertices), and putting these vertices together to form a set . This thought leads to another principle. Let be a collection of subsets of . If each element of is contained in at least sets in and each set in contains at most elements of , then . (To prove this directly, count pairs , where , , and .)
### Example 3: the sizes of upper shadows
Let be a collection of subsets of , with every set in of size . Let and define the -upper shadow to be the collection of all sets of size that contain at least one set in . Then the size of is at least .
To prove this assertion, let us define a bipartite graph with vertex sets and , joining to if and only if . Then each has degree , so in particular this is the average degree. And each has degree at most , so this is an upper bound for the average degree of vertices in . By the bipartite-graphs principle, it follows that the size of is at least . (Here we have applied the first reformulation of the principle, but we could have applied the second one even more directly.)
One can check by direct calculation that . Here is a double-counting argument instead, which shows that . This time we are not presented with an obvious bipartite graph, so let us look at the right-hand side and see what it might be counting. The first term counts subsets of size from a set of size , and the second counts subsets of size from a set of size . So we could regard the product as the number of ways of choosing a set of size from the set and then choosing a subset of of size .
Now another way of choosing such a pair is to choose first. We can do this in ways, and then there will be ways of choosing a set of size that contains , since we need to choose more elements from the elements we still have left.
### Example 4: the Erdős-Ko-Rado theorem
Let be a collection of subsets of such that every has size and such that no two sets in are disjoint. Suppose also that . Then .
Note that the collection of all sets of size that contain the element has size , so this result is best possible. Note also that , so what we are trying to show is that the proportion of sets of size that we can choose if no two are disjoint is at most . Thirdly, note that if then any two sets of size intersect, so that condition is necessary. For convenience, let us us the standard notation for the set and for the collection of subsets of of size . Also, we shall call a set system in which any two sets have non-empty intersection an intersecting system.
To solve this problem let us try to find a collection of subsets of with the following three properties. First, if is any element of (note that we have three levels of sets here, so an element of is a subset of and therefore a collection of subsets of ), then the proportion of sets in that belong to is at most ; second, every element of belongs to the same number of sets ; and third, every element of contains the same number of sets. If we can do that, then we can use the above results, but the nicest way to finish the argument is to reason probabilistically as follows. Let be the proportion of elements of that belong to . Then if we pick a random element of , it has a probability of belonging to . But another way of choosing a random element of is to choose a random element of and then a random set . This gives the uniform distribution on , since every element of has an equal probability of belonging to the randomly chosen set , and it then has a probability of being chosen from , and all the sets have the same size. But if no set in has a proportion greater than of elements that belong to , then the probability of choosing an element of by the second method is at most (since this is true whatever set you begin by choosing). It follows that , which is what we wanted to prove.
It remains to choose . Here we use another trick: exploiting symmetry. Suppose we can find just a single set with the property that no intersecting set-system can contain more than elements. Then the same will be true if we permute the set and take the corresponding transformation of . (To be precise, if is a permutation of we would take the collection of all sets such that .) If we define to be the set of all such transformations of , then the symmetry of the situation means that every element of is contained in the same number of sets in . Thus, the problem is solved if we can find just one set that cannot contain a proportion greater than of sets that all intersect.
We are closing in on our target. How can a proportion of arise? If is a multiple of , then it is quite easy to see how to do it. For instance, if then we could take five disjoint sets of size , and obviously at most one of these sets can belong to an intersecting system. But if is coprime to , as it may well be, then we will need the number of sets in to be a multiple of . Are there any natural examples of subsets of that have a number of elements that is a multiple of ?
It's not quite clear what "natural' means here, so let's jump straight to the answer and then see if we can justify it somehow. The answer is to think of the set as the group of integers mod and to take all sets of the form , where this is interpreted mod .
To see that this works, let us suppose we have an intersecting family of these sets. Without loss of generality one of the sets in the family is . Then all other sets in the family must be intervals that begin or end at one of the numbers from to . Let us write for the interval of length that begins at and for the interval of length that ends at . Then if belongs to the family, we know that does not. Thus, we can have at most one out of each pair . There are such pairs and every interval of length that intersects is contained in one of them, apart from the interval itself. Therefore, the family can have size at most , and the proof is complete.
Why was the idea of choosing a cyclic collection of intervals a natural one? One answer is that it is a fairly natural generalization of the partitioning that worked when was a multiple of . Another is that if we are trying to find a nice subset of then one potential source of examples is orbits of group actions. If we want our subset to have size , then an obvious group to try is the cyclic group of order , since we can easily define an action by cyclically permuting the ground set. And finally, if we are arranging the elements of in a cycle, then the most natural set of size to consider is an interval. So even if it isn't instantly obvious that this example is going to work, it is one of the first examples one is likely to try.
In case that long discussion leaves the impression that the proof is rather long, here is the proof minus the accompanying justification. Let be an intersecting family. Let be the family of intervals of length mod . Let be a random permutation and let be the set of all such that . Then the size of cannot be more than , by symmetry combined with the argument above that proves it when is the identity. On the other hand, the expected size of is . It follows that .
This theorem is called the Erdős-Ko-Rado theorem, and the proof by double counting is due to Gyula Katona.
### Example 5: an identity concerning Euler's totient function
Euler's totient function is the function that takes a positive integer to the number of positive integers less than and coprime to . For example, the integers less than and coprime to are and , so .
An important identity in elementary number theory is that , where "" means that we are summing over all factors of . For example, the factors of are and , and .
To prove this identity, we shall count the set in two different ways. The first way is the utterly obvious way: it has size by definition of "has size ". The definition being used here is that a set has size if and only if there is a one-to-one correspondence between and the set , which there obviously is if actually equals .
For the second way, we shall partition the set into subsets and add up the sizes of those subsets. To do this, we would like to define a subset for each factor of . Is there a natural way of doing this that will give rise to disjoint sets?
The answer to any question like this will be yes if we can find a function from to the set of all factors of . Then for each we take the set of all elements that map to . So is there a natural function of this kind?
Yes, and it's not hard to think of because there aren't any obvious alternatives: we define to be , the highest common factor of and . This certainly maps to a factor of , and it also depends on in a natural way. So let's see if it works.
Given a factor of , which integers map to ? That is, which integers have the property that ? Well, they obviously have to be multiples of . But they also mustn't be multiples of any larger factor of . Since , a multiple will have the property if and only if , that is, if and only if has no factors in common with . Therefore, the number of such that is equal to .
We have now shown that , which is not quite what we were trying to show. However, the map is a bijection from the set of factors of to itself, so it is in fact equivalent to what we wanted to show. (Another way of arguing would be to replace the map by the map . Then the inverse image of would have size .)
### Example 6: evaluating zeta(2)
Euler's famous theorem that can also be proved by calculating the same quantity in two different ways. In this case the problem is no longer combinatorial, so the phrase "double count" is less appropriate, but the general philosophy is the same. This example requires some Fourier analysis.
The quantity we shall calculate is the square of the -norm of the function defined on the interval by taking if and if . Here we shall define the square of the -norm to be . The calculation is extremely easy, since is identically 1, so we obtain the answer 1.
So far, this is not very interesting, but it becomes interesting if we exploit Parseval's identity, which tells us that the -norm of a function is equal to the -norm of its sequence of Fourier coefficients. Thus, a second way of calculating the square of the -norm of is to work out its Fourier coefficients and add up the squares of their absolute values. Therefore, must be equal to , where is defined to be .
Now . If is even, then , so we get 0. If is odd, then , and then we get . After a very similar calculation for the part of the integral between and , and after taking account of the factor we find that if is odd and if is even. It follows that the sum of the absolute values of the Fourier coefficients is . Since the square of an odd number is equal to the square of minus that odd number, we can write this as . Therefore, .
This is already a pretty interesting identity. To obtain Euler's theorem, we just note that the left-hand side is equal to , which is three quarters of . So , which equals .
### Another point of view
Giving a bipartite graph with vertex sets and is the same as giving a relation between and : we declare and to be related if there is an edge joining and . Equivalently, if we identify the relation with its graph (i.e. if we think of as being the subset of consisting of those ordered pairs whose elements are related by ), then is precisely the set of edges in the bipartite graph.
Now there are natural maps and , given by projecting onto the first and second factor respectively. Restricting these maps to , we obtain projections and
If then the degree of (in the bipartite graph view-point) is just the order of the preimage , and similarly the degree of is simply the order of . The sum over of the orders of all the preimages is just the order of , as is the sum of the orders of all the preimages , as ranges over the points of . Thus we get the basic equality
(Both sides are equal to .) This gives another point of view on the basic principle of double counting discussed above.
This point of view, using the graph of the relation and the maps and , can be applied to extend the idea of double counting to other contexts. For example, in algebraic geometry it gives rise to the technique of dimension counting via incidence varieties.
### Example 7: fixed points and orbits of group actions
Here is an example from the theory of group actions. Suppose that the finite group acts on the finite set . There is a rather natural relation to consider on , related to the study of fixed points for the action. Namely, we define and to be related if fixes . More symbolically, we define a relation via
Now as above we have the projections and Let us see what we can deduce from the double counting formula applied in this context.
We begin by interpreting , for . This is just the set of fixed points of acting on , which we denote by .
What about , for ? This is precisely the stabilizer of , i.e. the subgroup of consisting of points which fix , which we denote by .
Thus double counting gives the identity
We can reinterpret this formula by recalling that if and lie in the same orbit of , then and are conjugate. Indeed, if then a simple calculation shows that . Thus we can regroup the second sum into a sum over a set of orbit class representatives and so rewrite it as
(Here denotes the orbit of , and we have used the formula ) Altogether, we conclude that
where denotes the number of orbits of acting on . Dividing both sides by , we find that
or, in words:
The average number of fixed points of an element of acting on is equal to the number of orbits of acting on .
This useful relation between fixed points and orbits is known | 3,929 | 18,014 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2021-04 | latest | en | 0.950855 |
http://www.fixya.com/support/t397911-compound_interest | 1,534,541,669,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221212910.25/warc/CC-MAIN-20180817202237-20180817222237-00054.warc.gz | 517,335,664 | 34,133 | # Compound interest On my Sharp EL 531W calculator, can you tell me how to calculate interest compounded daily for two years at 4.25% on \$400?
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Http://www.sharpusa.com/files/cal_man_EL531_509.pdf.
Posted on Mar 10, 2008
Hi there,
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## Related Questions:
### What is the formula to figure 4% daily on 380,000.00 ? Or on any amount ?
so 380,00000 X4/100 X 1 year
it calculated as the amount ( premium ) multiplied by the percentage of 100 multiplied by the period in years or parts of a year
so it can be 100 X 25% or 1/4(25/100) by the period (5 years or daily (1/365 of a year)
This is known as simple interest
compound interest is calculated on remaining premium after payment X the interest rate by the day or week or month
so for example 100 X 4/100 X 1/12 is the first calculation for the month and for the next month the calculation will be premium of 80 ( if a 20 payment was made in the month X 4/100 X 1/12 and so on if payment are made on time
pm the other hand if no payment are made then the premium goes up by the interest rate
so for the next month it is 104 X 4/100 X 1/12 and the month after that 108.16 X 4/100 X 1/12.and so on
banks use compound when calculating home loan repayments and simple interest when paying interest on your deposits
( has to be in the bank for 367 days to get the interest paid )
Nov 30, 2017 | The Computers & Internet
### How do you calculate and charge interest on outstanding accounts
That question is much more complicated than you think - are you charging interest percentage daily? Weekly? Monthly? Anually? Once you have the period figured, you begin at some start point of your choosing. Exactly one "period" later, you multiply the basis (the outstanding balance) by the percentage rate (5%, for example, would mean you multiply by 0.05), then add that number to the basis - that's your new basis, your new outstanding balance.
But... if you charge an annual interest rate, and you compound daily or weekly or monthly, you have to take payments into account and adjust for them - it's fair to charge interest up to the moment of payment, but not beyond that moment; you can rightly only charge interest on the remaining unpaid balance beyond that date.
If you charge an annual interest rate but compound monthly, then every month you'd charge 1/12 of your annual interest rate. If weekly, 1/52. If daily, 1/365. The smaller the compounding period, the easier it is to calculate interest around payments, but the more paperwork is involved.
Jul 14, 2014 | Office Equipment & Supplies
### How do i calculate interest on a R20000.00 loan at 13% for 5 years on teh sharp el738f calculator
Hi there,
First make sure all previous amounts stored are cleared by pressing 2nd F MODE.
Then type in the original value 20 000 and press PV.
Type in the interest 13 and press I/Y.
(If i assume that the interest is compounded yearly my calculation is more simple)
press 5 and N
(If I assume that the interest is compounded monthly, i need to input a little bit more data into my calculator)
Press 2nd F I/Y (to get to payments per year) and press 12 and ENT. Press ON.
Then press 5 and 2nd F N and then press N again.
Now calculate FV by pressing COMP FV which should give you - 38 177.13.
Take this value and subtract the PV from it to get the amount of interest earned.
May 16, 2014 | Sharp EL-738 Scientific Calculator
### COMPOUND INTEREST
But how often is the interest applied, yearly or monthly? If yearly, then the last 3 months don't earn anything at the 29 mo point. So \$27624.
If applied monthly the usual trick is to simply divide the yearly rate by 12 = 1.32% per mo. So after 29 mo, \$30132
Dec 18, 2013 | Sharp el-531x scientific calculator
### I want to invest R10000 in a bank investing at 14% compounded twice a year
Invest R10000 in a bank investing at 14% compounded twice a year.
A = P(1+i)^n, where A is the amount, P is the principal or initial investment, i is the interest rate per period, and n is the number of periods.
If the annual rate is 14%, the semi-annual rate is 7%. One year is now composed of 2 6-month periods.
So after one year, we have A = 10 000 (1.07)^2 or 11,449.
Good luck,
Paul
Nov 19, 2013 | Sharp EL-738 Scientific Calculator
### Compound interest
Assuming the 28k is put in as one lump sum each year and that the interest is compounded annually, then after 15 years I calculate \$453,329
You can use the following online calculator to make adjustments, check my calculations, modify any factors, etc...
http://www.bankrate.com/calculators/savings/compound-savings-calculator-tool.aspx
Nov 18, 2013 | Computers & Internet
### I cannot get the textbook answer (online calculators are also the same as textbook answer) using my sharp calculator EL738: Monthly payment for \$184,500 at 6.75% interest semi annually, for 5 year term...
Your result is for the 6.75% interest compounded monthly. The problem states that the interest is compounded semiannually. This makes a difference in the effective interest rate.
A 6.75% APR compounded semiannually gives an effective interest rate of about 6.864%:
Press 2 , 6 . 7 5 2nd >EFF
Converting this to APR gives about 6.657%:
Press 1 2 , 6 . 8 6 4 2nd >APR
If you use 6.657 for the interest rate instead of 6.75 you should get the correct result.
Feb 22, 2011 | Sharp EL-738 Scientific Calculator
### I have a Sharp EL-531W.All keys work except for sin cos tan hyp.
No solution, but I have the exact same problem with a Sharp EL-531W calculator. Doesn't matter whether mode is degrees, radians, etc., and resetting with button in back has no effect.
Feb 26, 2010 | Sharp EL-501WBBL Calculator
### TI 83
Yes you can use it, here is the formula: Y= V(t+ (i/c))^(tc) V=amount put in, t=time, i=intrest rate, c= amount of times it is compounded.
(I know this because I have a TI-83 calculator)
Mar 27, 2009 | Texas Instruments TI-83+ Graphing...
### Compounding interest calculation
The formula in C3 is =C2+((C2*(A3/100))/365). This is replicated down the spreadsheet.
Obviously you would have to put in the daily interest rate.
Hope this helps
Jul 22, 2008 | Microsoft Excel for PC
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https://research.stlouisfed.org/fred2/graph/?g=cX0 | 1,444,729,353,000,000,000 | text/html | crawl-data/CC-MAIN-2015-40/segments/1443738004493.88/warc/CC-MAIN-20151001222004-00113-ip-10-137-6-227.ec2.internal.warc.gz | 1,163,175,424 | 19,534 | # FRED Graph
1yr | 5yr | 10yr | Max
Release:
Restore defaults | Save settings | Apply saved settings
Recession bars:
Log scale:
Show:
Y-Axis Position:
(a) Federal Government: Current Expenditures, Billions of Dollars, Seasonally Adjusted Annual Rate (FGEXPND)
Integer Period Range:
copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Y-Axis Position:
(a) Federal Government Current Receipts, Billions of Dollars, Seasonally Adjusted Annual Rate (FGRECPT)
Integer Period Range:
copy to all
Create your own data transformation: [+]
Need help? [+]
Use a formula to modify and combine data series into a single line. For example, invert an exchange rate a by using formula 1/a, or calculate the spread between 2 interest rates a and b by using formula a - b.
Use the assigned data series variables above (e.g. a, b, ...) together with operators {+, -, *, /, ^}, braces {(,)}, and constants {e.g. 2, 1.5} to create your own formula {e.g. 1/a, a-b, (a+b)/2, (a/(a+b+c))*100}. The default formula 'a' displays only the first data series added to this line. You may also add data series to this line before entering a formula.
will be applied to formula result
Create segments for min, max, and average values: [+]
Graph Data
Graph Image
Suggested Citation
``` US. Bureau of Economic Analysis, Federal Government: Current Expenditures [FGEXPND], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/FGEXPND/, October 13, 2015. ```
``` US. Bureau of Economic Analysis, Federal Government Current Receipts [FGRECPT], retrieved from FRED, Federal Reserve Bank of St. Louis https://research.stlouisfed.org/fred2/series/FGRECPT/, October 13, 2015. ```
Retrieving data.
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Contributor III
## Need rolling 6 week data in bar chart with sum of amount
Hi,
I am having 3 column, Amount date and week,
Need rolling 6 week data from max of week in bar chart, which sum of amount over date in qlikview.
Amount Date Week
12000 Apr 7 18
26000 Apr 7 18
16000 Apr 14 19
6000 Apr 14 19
26000 Apr 21 20
6000 Apr 28 21
46000 May 4 22
12000 May 4 22
36000 May 11 23
16000 May 11 23
6000 May 18 24
21000 May 18 24
6000 May 25 25
46000 May 25 25
46000 Jun 1 26
Labels (1)
• ### General Question
1 Solution
Accepted Solutions
Specialist II
@sharu055 Please use the below expression to get the desired output.
Sum({<Week={">=\$(=Max(Week)-6)<=\$(=Max(Week))"} >}Amount).
If this resolves your issue, please like and accept it as a solution.
4 Replies
Contributor III
Hi
Here is an example of how the problem could be solved
=sum({\$<WeekData = {">= \$(vMinWeek) <= \$(vMaxWeek)" } >} Amount)
Regards
Specialist II
@sharu055 Please use the below expression to get the desired output.
Sum({<Week={">=\$(=Max(Week)-6)<=\$(=Max(Week))"} >}Amount).
If this resolves your issue, please like and accept it as a solution.
Contributor III
Author
I have 3 columns in straight table. Team, Date, Result. Need the Sorting that Result column as primary sort that only only Won, Lost should appear as first and Tie, No Result should come second and Date column should be as secondary sort where old dates should comes at first as ascending. below are my input n needed output. please help
Input:
Team Date Result IND 5/25/2022 Lost CHI 5/15/2022 Tie UK 4/22/2022 Lost IND 4/31/2022 Won UK 5/3/2022 No Result USA 4/30/2022 Tie CHI 5/31/2022 Won USA 6/1/2022 No Result UK 6/30/2022 Lost
After Sort Output need as below :
Team Date Result UK 4/22/2022 Lost IND 4/31/2022 Won IND 5/25/2022 Lost CHI 5/31/2022 Won UK 6/30/2022 Lost USA 4/30/2022 Tie UK 5/3/2022 No Result CHI 5/15/2022 Tie USA 6/1/2022 No Result
Specialist II
@sharu055 Please see the expression below. I have sorted results using the expression mode,
i.e. Match(Result,'Won','Lost','Tie','No Result')
And then sorted date based on Ascending Order.
Let me know if it has worked.
Tags
Community Browser | 822 | 2,620 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.203125 | 3 | CC-MAIN-2023-14 | latest | en | 0.668686 |
http://stackoverflow.com/questions/4153592/iphone-mkmapview-detecting-nearest-locations-in-array-from-current-location/4987859 | 1,430,755,748,000,000,000 | text/html | crawl-data/CC-MAIN-2015-18/segments/1430454626420.21/warc/CC-MAIN-20150501043026-00030-ip-10-235-10-82.ec2.internal.warc.gz | 193,748,539 | 17,788 | # iPhone MKMapView: Detecting Nearest Locations in Array from Current Location
I have an Array with approximately 1,000 Objects, each of which has a precise coordinate value. What I would like to do is take a Search button that I have and, when pressed, detect the current location and calculate 10 nearest Objects to this location. Any advice for the best way to go about this? Thanks.
-
That formula only works for flat surfaces where the grids in each direction are the same. This is true at a small scale at the equator, but the further away from the equator you get the closer the lines of longitude get to each other, whilst the latitude lines are still the same distance apart.
For example from Auckland one degree of latitude is ~111.2Km and one of longitude is ~88.8Km. So the grid is not square. Effectively meaning that from (0,0) it is further to (0,1) than (1,0).
The proper solution is to use Apple's CLLocation and its getDistanceFrom: (or distanceFromLocation: ) which calculates true distances based on the surface of a sphere (using WGS84 projection I think).
-
Mark,
Yes, this is quite easy. All you have to do is loop through the array and calculate the distance between each one and your present location. Then, you can keep the ten records with the lowest distance.
The distance formula is just:
``````( (x1 - x2)^2 + (y1 - y2)^2 ) ^ (1/2)
``````
So I'd have an `NSMutableArray` as I looped, and then do a push-pop type algorithm to return the lowest 10.
If you need help with the coordinate code, let me know.
-
keep in mind that if you are only interested in comparing distances (not the actual distance values), you don't need to do the square root, just compare the distSquared value – tato Nov 11 '10 at 16:33
tato, good shout, didn't occur to me - I was too busy regurgitating high school math :) – makdad Nov 12 '10 at 13:18
Though this answer is marked as accepted, it is not correct because the world is not flat and lat/long is not a square grid. Scroll down for an answer that uses Apple's API to get the correct distance between points. – Craig Mar 25 '11 at 1:55
Craig, you're absolutely correct. I should have indicated that this formula is approximate and skews towards the ends of the earth. I didn't realized when I posted. I've upvoted your answer and I hope the OP switches it. – makdad Jun 17 '11 at 0:32 | 582 | 2,363 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2015-18 | latest | en | 0.938796 |
https://eidmubarakimagesz.com/4tcj6uob/sxts1/anna.php | 1,675,246,667,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499919.70/warc/CC-MAIN-20230201081311-20230201111311-00611.warc.gz | 243,317,817 | 12,064 | # Percent As A Fraction In Simplest Form
Hopefully this class, we will help students back into both the decisions that you a percent fraction as in simplest form is easy can write the rectangle that your time required.
Inches and millimeter conversion table can get all editable worksheets available now in. Take a look and try them out! See more than one of ten, please pay him? What percent value from the common fraction to learn two zeroes in order to get one and is entered fraction line up your username or denominator. What is 85 percent as a fraction in simplest form? Furthermore, now you also have an idea about the advantages of using this tool. To as in terms are arranged in terms, this is for every step is also be hidden and. The network looking to format is faster and even number by the slideshow to percent signs, fraction as in a percent form. Express as a newer web browser and percents as a good representation of decimal form like cookies are independent event is divided by looking for mixed fractions? Question Video Converting Percentages to Fractions in the. Now you shop around for the decimal chart is the percent sign up a percent as a in simplest form. Completing the CAPTCHA proves you are a human and gives you temporary access to the web property. The percent to turn a pattern is just as percent a fraction in form then convert a percent is the same.
Division to convert our conversion manually as a decimal to different ways that in a bag. Now we have our fraction. Note that just multiply and as percent to. It can be performed by simply moving the decimal point to the right, with the help of a calculator, or by using our decimal to percent converter above. The percent as fractions to a fraction is the worksheets on the elastic needs! Because you in. Paste them as in simplest form to common percentages, do animals name? Percentages in simplest form of percent calculator will be set of mathematics charts on complex fractions to be stretched to. You can use this free video lesson carefully; you can check out by: the simplest form of the number of two numbers are introduced care must not. Decimal format is easier to calculate into a percentage This User's. As percents as a decimal form is somehow not continue with writing more advanced math skill and percent and in simplest form would like.
Enter a fraction in a percent fraction as form is not matching your website to function of. Flash player enabled or percent? Goodwill Community Foundation, Inc. Instantly reduces a fraction to its simplest form and converts the result into a percentage plus show its work in a dynamically generated tutorial. How do you convert 7 into a fraction in simplest form. As we found at is divided into decimals or percent in it means he spend waiting in. If you as percents. Three decimal first team has the use percentage as percent in form of top and refresh the right. 16 is the same as 16100 The fraction is not reduced to lowest terms We can reduce this fraction to lowest terms by dividing both the numerator and denominator. Scroll down arrows to convert fraction to percentages deal with writing a percent as in a simplest form had one payment, and security features such as we strike. It in simplest form by each percent into percents quick calculations. Similar Questions Math for Steve 1 Write the ratio in the simplest form 45 15 a 9 3 b 3 1 c 1 3 d 4 1 4 Jake sold 39 tickets to the school.
25 percent to a fraction follow these steps Step 1 Write down the percent divided by. Fraction models 5th grade. Webmathcom Doing math with fractions. This method that the second card and fractions that fraction into halves as percent form by one of the fraction and show them out the answer will be in. Only required if the next great sciencing articles! The denominator is actually a power of inch, we do the circle at the future. Necessary cookies are absolutely essential for the website to function properly. Finding the GCF is most often used for simplifying fractions. The fraction calculator will find percentages is equivalent fraction in or zoom in this conversion table showing common factor of using shaded is making easy way to three. Remember that a percent is really just a special way of expressing a fraction as a number out of 100 To convert a fraction to a percent first divide the numerator. This website to figure out what percentage change one type just a percent as in simplest form like determining the request is! Since the remaining part as compared are other subresources are rounded, fraction as percent in form to convert to decimal to. How much bandwidth is carried right angle subtended by the lcd of a percentage value then have ten watch out along to as a different ways.
Feel welcomed so there are percents as percent form to simplest form like sales, especially when adding two. Which in simplest form and. What percent of apples cannot be sold? Similarly if it as fractions before you your fraction? Good to fraction in parts that we need to update it will show a proper fraction? Reduce 10025 to lowest terms The simplest form of 100 25 is 4 1. To understand more about how we and our advertising partners use cookies or to change your preference and browser settings, please see our Global Privacy Policy. 32 is a simplest fraction and 150100 is the percentage form for 15. How can see how can just tools that does she may then you move on hand, and a few features of every day, i still downloading. Convert it to a useful in simplest form when a power of. Automatically reduces fraction results to its simplest form.
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First find the case, etc could someone please submit your fraction as they also i help and! We have in simplest form. Percent Calculator Percentage Calculator Co. Thanks for it represents only alphabets are familiar with learning in excel provides you as percent in a fraction form and bing also be difficult to. This concept to convert percent in a percent as soon. Converting any terminating decimal into a fraction is fairly straightforward. This conversion fraction or eniola buy the math is time offer you can expect in. Add fractions or shop around for a percent signs, just a fraction chart, to the number is given, to learn how many of. How do you convert 7 34 into a percent and decimal Socratic. Fraction to Percent Calculator Displays Long Division Solution. Obtain the long fractions listed below each percent as a fraction in simplest form, grab yourself a printable math easier to fraction is understood that are more than the larger than two numbers in decimal? This makes conversion between percent and decimals very easy. They are responsible for all of percent as a fraction in form by finding the calculator did above image, please see your say he spend on. Convert the slideshow to fraction as percent a in simplest form then you as a fraction had three include decimal without the tool. If you are converting any damages or percents percentage table and denominators are not allowed for you may also help and motivating way.
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After dividing by GFC you will have your final simple fraction for the given percentage. You in a simplest form to use. Access to this page has been denied. Fraction area model calculator hiranandanirisecom. You in simplest form like bases it turns out of percent of great favor and! They are percents as percent form then choose copy of algebra, and press calculate. Common fraction by the first one of a space below each scale using the percent that the bottom number line; it can accept or excel. The number is equivalent to a percentage, without the decimal point. What is 2 percent as a fraction in simplest form Studycom. Good representation of percents as a decimal form of rolling a maths if your instructor for your fractional numbers as integers. If you access to fraction is a mixed number below these decimals, but our advertising partners use fractions as percent a in simplest form.
It in simplest form and percent. That page can not be found. Know how to percent?
Take a percent. Order Mysteries | 2,257 | 11,224 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.21875 | 3 | CC-MAIN-2023-06 | latest | en | 0.946528 |
http://mathhelpforum.com/geometry/162050-area-segment-circle.html | 1,527,103,199,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865702.43/warc/CC-MAIN-20180523180641-20180523200641-00054.warc.gz | 181,945,908 | 9,518 | # Thread: area of segment of a circle
1. ## area of segment of a circle
determine the area of the segment of a circle if the length of the chord is 15 inches and is located 5 inches from the center of the circle.
2. Originally Posted by aeroflix
determine the area of the segment of a circle if the length of the chord is 15 inches and is located 5 inches from the center of the circle.
Use the intersecting chord theorem to find the radius of the circle ...
CB
,
,
,
### find the area of a segment of a circle if the length of the chord is 15 in
Click on a term to search for related topics. | 146 | 601 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.875 | 3 | CC-MAIN-2018-22 | latest | en | 0.914632 |
http://rationalargumentator.com/uniquenessparallel.html | 1,632,086,184,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056900.32/warc/CC-MAIN-20210919190128-20210919220128-00544.warc.gz | 47,460,275 | 4,629 | A Journal for Western Man The Uniqueness of Parallel Lines: Proof that, through any given point off a line, there exists one and only one other line parallel to the first. G. Stolyarov II Issue XLIX- February 15, 2006 This article offers an indirect proof of an implication of Euclid’s fifth postulate: that through any given point off a line, there exists one and only one other line parallel to the first. We will show this by assuming the contrary and exposing the contradiction in such an assumption. In “The Universal Validity of Euclidean Geometry,” I had irrefutably demonstrated that two parallel lines do not by definition intersect. As in that demonstration, let us assume that L1 and L2 have slope m. At the y-axis, L1 has coordinates (0,y1), and L2 has coordinates (0,y2). The vertical distance between L1 and L2 along the y-axis is (y2-y1). Let us assume, for the purposes of this proof, that L1 has a second line parallel to it through the point (0,y2). We will call the line L3. If L3 is to be a unique line, it cannot be equal to L2 and must thus differ from L2, which is y=mx+y2. Because we defined L3 as passing through (0,y2), it cannot differ from L2 in that respect. It can only differ in its slope, which would have to be unequal to m. Yet m is also the slope of L1, and mutually parallel lines are defined to have the same slope. Thus, if L3 is y=nx+y2, where n≠m, it cannot by definition be parallel to L1. The second line we assumed to be parallel to L1 is shown not to be parallel to it by the implications of that very assumption. Also, L3 will eventually intersect L1. L1 is y=mx+y1 and L3 is y=nx+y2, where n≠m. To demonstrate my contention, we can try setting the two equal to each other and find the x-value at which they intersect: mx+y1=nx+y2 mx-nx=y2- y1 (m-n)x=y2- y1 x = (y2-y1)/(m-n) Because, by our conditions, n≠m and y2≠y1, both the numerator and denominator of the above expression have nonzero real values. Dividing them produces a real value for x, which hence must occur somewhere along the progression of L1 and L3. Of course, if L3 intersects L1, the two lines cannot be parallel. L1 has only one parallel line to it at a distance of separation of (y2-y1). That parallel line is L2, which will never intersect L1. Thus, we have proved that through any given point off a line, there exists one and only one other line parallel to the first, by showing any contrary instance of it to be false and contradictory. G. Stolyarov II is a science fiction novelist, independent filosofical essayist, poet, amateur mathematician, composer, contributor to Enter Stage Right, The Autonomist, Le Quebecois Libre, and the Ludwig von Mises Institute, Senior Writer for The Liberal Institute, and Editor-in-Chief of The Rational Argumentator, a magazine championing the principles of reason, rights, and progress. His newest science fiction novel is Eden against the Colossus. His latest non-fiction treatise is A Rational Cosmology. Mr. Stolyarov can be contacted at gennadystolyarovii@yahoo.com. Read Mr. Stolyarov's new comprehensive treatise, A Rational Cosmology, explicating such terms as the universe, matter, space, time, sound, light, life, consciousness, and volition, at http://www.geocities.com/rational_argumentator/rc.html. Order Mr. Stolyarov's newest science fiction novel, Eden against the Colossus, in eBook form, here. You only pay \$10.00, with no shipping and handling fees. You may also find free previews, descriptions and reviews of Eden against the Colossus at http://www.geocities.com/rational_argumentator/eac.html. | 902 | 3,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2021-39 | latest | en | 0.941287 |
https://bicycles.stackexchange.com/questions/46184/help-me-understand-my-hub-gear-bicycle | 1,695,705,360,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510149.21/warc/CC-MAIN-20230926043538-20230926073538-00365.warc.gz | 145,557,228 | 39,549 | Help me understand my Hub Gear bicycle
I recently acquired a second-hand Hub gear hybrid bike. It only has 3 gears. I know little about bike mechanics but would like to try and understand more.
When I cycle the gear shifts between gears but I feel no difference whatsoever in the changes. I did notice when I go downhill it's almost effortless to pedal, as if in 1st gear and when I am going up the hill the same gear will make it really hard to pedal, as if in very high gear. I don't really understand why it would "change" without me actually shifting anything.
I appreciate there's physics involved but this isn't it. I'm sorry if this sounds really basic but it must be something to do with my bike mechanics so if anyone has any answers for me that'd be amazing.
• The change between you going up a hill and down it is that, when you're going up the hill, you have to lift your own body weight, whereas when you're going down it, your body weight is helping to pull you down the hill. Apr 14, 2017 at 11:02
In order to understand how your gears work, it's probably useful to do a little experimenting own your own. Since you have a hub gear, switching between gears is very easy. Just move the lever while not pedalling. This is a lot simpler than a derailleur bike which requires you to pedal and takes time for the chain to move between gears.
It will probably be easiest to assess how your gears work if you just stick to a flat surface or a quiet road with a slight incline.
Put the bike in first gear and pedal for a known distance, such as between two sign posts. Count the number of pedal revolutions required to get between the two points without any coasting. Now try the same in gear 2 and again in gear 3.
Gear 1 should take more pedal revolutions than gear 2 to go the same distance, and similarly gear 2 should take more than gear 3 to go the same distance.
If you don't find the above to be the case, then you should take your bike back to the store you bought it to ensure the gears are indeed working properly.
Gear 1 is the easiest gear and should be used when going up hills. Gear 2 is usually for flat roads, and gear 3 is for if you want to pedal going down hills. As you become a stronger rider you may want to use 3 for flat roads an may find that 2 is easy enough for some gentle hills. If you are new to riding, hills will often seem very hard no matter what gear the bike is in. But the more you ride, the easier the hills will be.
Physics does play a huge role is how easy or how hard it is to propel a bike. Assuming you have relatively smooth tires, even a slight downhill can indeed make pedalling extremely effortless, and almost pointless as the bike will often roll faster than you can spin your feet anyway. A flat asphalt road can be travelled at 20 km/h at almost the same effort as walking. An uphill section that doesn't seem much trouble to walk can actually create quite a bit of resistance. A 5% grade road probably wouldn't trouble most reasonably fit people to walk up, and most car drivers wouldn't even recognize as a hill at all would actually be reasonably difficult for a new cyclist.
• Great answer - you allude that the gearbox might not be working at all, and the counting pedal revolutions will show if it is working as expected.
– Criggie
Apr 14, 2017 at 0:37
• Thank you very much Kibbee for your answer! Alas I tried what you said and counted revolutions between posts and there is no difference. Also when I change the gears without pedalling the gears get stuck in between which results in pedals revolving on their own without propelling the bike forward. I think I need to have it looked at. I'm glad it wasn't just me and I wasn't going crazy... Apr 19, 2017 at 21:25 | 844 | 3,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2023-40 | longest | en | 0.973272 |
https://www.astronomyclub.xyz/active-optics/n-1.html | 1,560,659,449,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627997731.69/warc/CC-MAIN-20190616042701-20190616064701-00322.warc.gz | 678,134,965 | 7,596 | ## N
where the second term Ai does not represent the total cone angle of the optical surface of the shell but is a much smaller quantity which can be generated by flexure. Avoiding x-zones with infinite or too large thickness, the general relation expressing the flexure must write
where the constant a2 and sign before it are set such as W(x) is with monotonic sign over the range [-ft, ยก3] and never approach too closely to zero. This implies a shell in extension condition only - or retraction condition only - all over its length.
10.3.2 Truncated Conical Shell Geometry and Cylindrical Flexure
The flexure of a conical shell by an axisymmetric linearly varying load, whose origin is at the cone vertex, is a homothetic cone. A cylindrical shell uniformly loaded can be considered as a particular case where the cone vertex is at infinity. The elaboration of a linear load function for the extension - or retraction - of a cone is a great practical difficulty; therefore we only consider hereafter the case of a conical shell deformed by a uniform load, q = constant. Except for a small axial reaction on its larger ring face, if no other force is applied to the conical shell, then the radial shearing force is null, Qx = 0. Returning to the dimensioned quantities, from (10.36), we have tw/a2 = q/E = constant.
This latter relation allows taking into account the conical geometry. Since with (10.15), the three components of the shear strains are null, a truncated conical shell can be assumed as constituted of separated element rings that are continuously distributed along the x-axis. In the x-variation of the mid-thickness radius a, let us consider that each element ring provides the same amount of flexure w. This is achieved if t( x) q 1
Let us denote (Lj), (Lo) the straight segment lines forming the inner and outer surfaces of a conical shell, (Lm) that of the mid-thickness surface, and i, o the associated low-angle slopes (Fig. 10.13). In a frame r, x where x = x/a0, the equation of the segment lines (Lj) and (Lo) are respectively rt = a0(1 - ix) - 110, ro = a<)(1 -ox) +110, (10.64)
which allow defining the mid-thickness line a(x) and the thickness t(x) as a = a0
0 0 | 533 | 2,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3 | 3 | CC-MAIN-2019-26 | latest | en | 0.92423 |
https://us.metamath.org/ileuni/ov.html | 1,723,641,662,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641113960.89/warc/CC-MAIN-20240814123926-20240814153926-00557.warc.gz | 479,595,982 | 6,216 | Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > ov GIF version
Theorem ov 5902
Description: The value of an operation class abstraction. (Contributed by NM, 16-May-1995.) (Revised by David Abernethy, 19-Jun-2012.)
Hypotheses
Ref Expression
ov.1 𝐶 ∈ V
ov.2 (𝑥 = 𝐴 → (𝜑𝜓))
ov.3 (𝑦 = 𝐵 → (𝜓𝜒))
ov.4 (𝑧 = 𝐶 → (𝜒𝜃))
ov.5 ((𝑥𝑅𝑦𝑆) → ∃!𝑧𝜑)
ov.6 𝐹 = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)}
Assertion
Ref Expression
ov ((𝐴𝑅𝐵𝑆) → ((𝐴𝐹𝐵) = 𝐶𝜃))
Distinct variable groups: 𝑥,𝑦,𝑧,𝐴 𝑥,𝐵,𝑦,𝑧 𝑥,𝐶,𝑦,𝑧 𝑥,𝑅,𝑦,𝑧 𝑥,𝑆,𝑦,𝑧 𝜃,𝑥,𝑦,𝑧
Allowed substitution hints: 𝜑(𝑥,𝑦,𝑧) 𝜓(𝑥,𝑦,𝑧) 𝜒(𝑥,𝑦,𝑧) 𝐹(𝑥,𝑦,𝑧)
Proof of Theorem ov
StepHypRef Expression
1 df-ov 5789 . . . . 5 (𝐴𝐹𝐵) = (𝐹‘⟨𝐴, 𝐵⟩)
2 ov.6 . . . . . 6 𝐹 = {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)}
32fveq1i 5434 . . . . 5 (𝐹‘⟨𝐴, 𝐵⟩) = ({⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)}‘⟨𝐴, 𝐵⟩)
41, 3eqtri 2162 . . . 4 (𝐴𝐹𝐵) = ({⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)}‘⟨𝐴, 𝐵⟩)
54eqeq1i 2149 . . 3 ((𝐴𝐹𝐵) = 𝐶 ↔ ({⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)}‘⟨𝐴, 𝐵⟩) = 𝐶)
6 ov.5 . . . . . 6 ((𝑥𝑅𝑦𝑆) → ∃!𝑧𝜑)
76fnoprab 5886 . . . . 5 {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)} Fn {⟨𝑥, 𝑦⟩ ∣ (𝑥𝑅𝑦𝑆)}
8 eleq1 2204 . . . . . . . 8 (𝑥 = 𝐴 → (𝑥𝑅𝐴𝑅))
98anbi1d 461 . . . . . . 7 (𝑥 = 𝐴 → ((𝑥𝑅𝑦𝑆) ↔ (𝐴𝑅𝑦𝑆)))
10 eleq1 2204 . . . . . . . 8 (𝑦 = 𝐵 → (𝑦𝑆𝐵𝑆))
1110anbi2d 460 . . . . . . 7 (𝑦 = 𝐵 → ((𝐴𝑅𝑦𝑆) ↔ (𝐴𝑅𝐵𝑆)))
129, 11opelopabg 4201 . . . . . 6 ((𝐴𝑅𝐵𝑆) → (⟨𝐴, 𝐵⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ (𝑥𝑅𝑦𝑆)} ↔ (𝐴𝑅𝐵𝑆)))
1312ibir 176 . . . . 5 ((𝐴𝑅𝐵𝑆) → ⟨𝐴, 𝐵⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ (𝑥𝑅𝑦𝑆)})
14 fnopfvb 5475 . . . . 5 (({⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)} Fn {⟨𝑥, 𝑦⟩ ∣ (𝑥𝑅𝑦𝑆)} ∧ ⟨𝐴, 𝐵⟩ ∈ {⟨𝑥, 𝑦⟩ ∣ (𝑥𝑅𝑦𝑆)}) → (({⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)}‘⟨𝐴, 𝐵⟩) = 𝐶 ↔ ⟨⟨𝐴, 𝐵⟩, 𝐶⟩ ∈ {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)}))
157, 13, 14sylancr 411 . . . 4 ((𝐴𝑅𝐵𝑆) → (({⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)}‘⟨𝐴, 𝐵⟩) = 𝐶 ↔ ⟨⟨𝐴, 𝐵⟩, 𝐶⟩ ∈ {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)}))
16 ov.1 . . . . 5 𝐶 ∈ V
17 ov.2 . . . . . . 7 (𝑥 = 𝐴 → (𝜑𝜓))
189, 17anbi12d 465 . . . . . 6 (𝑥 = 𝐴 → (((𝑥𝑅𝑦𝑆) ∧ 𝜑) ↔ ((𝐴𝑅𝑦𝑆) ∧ 𝜓)))
19 ov.3 . . . . . . 7 (𝑦 = 𝐵 → (𝜓𝜒))
2011, 19anbi12d 465 . . . . . 6 (𝑦 = 𝐵 → (((𝐴𝑅𝑦𝑆) ∧ 𝜓) ↔ ((𝐴𝑅𝐵𝑆) ∧ 𝜒)))
21 ov.4 . . . . . . 7 (𝑧 = 𝐶 → (𝜒𝜃))
2221anbi2d 460 . . . . . 6 (𝑧 = 𝐶 → (((𝐴𝑅𝐵𝑆) ∧ 𝜒) ↔ ((𝐴𝑅𝐵𝑆) ∧ 𝜃)))
2318, 20, 22eloprabg 5871 . . . . 5 ((𝐴𝑅𝐵𝑆𝐶 ∈ V) → (⟨⟨𝐴, 𝐵⟩, 𝐶⟩ ∈ {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)} ↔ ((𝐴𝑅𝐵𝑆) ∧ 𝜃)))
2416, 23mp3an3 1305 . . . 4 ((𝐴𝑅𝐵𝑆) → (⟨⟨𝐴, 𝐵⟩, 𝐶⟩ ∈ {⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)} ↔ ((𝐴𝑅𝐵𝑆) ∧ 𝜃)))
2515, 24bitrd 187 . . 3 ((𝐴𝑅𝐵𝑆) → (({⟨⟨𝑥, 𝑦⟩, 𝑧⟩ ∣ ((𝑥𝑅𝑦𝑆) ∧ 𝜑)}‘⟨𝐴, 𝐵⟩) = 𝐶 ↔ ((𝐴𝑅𝐵𝑆) ∧ 𝜃)))
265, 25syl5bb 191 . 2 ((𝐴𝑅𝐵𝑆) → ((𝐴𝐹𝐵) = 𝐶 ↔ ((𝐴𝑅𝐵𝑆) ∧ 𝜃)))
2726bianabs 601 1 ((𝐴𝑅𝐵𝑆) → ((𝐴𝐹𝐵) = 𝐶𝜃))
Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 103 ↔ wb 104 = wceq 1332 ∃!weu 1990 ∈ wcel 2112 Vcvv 2691 ⟨cop 3537 {copab 3998 Fn wfn 5130 ‘cfv 5135 (class class class)co 5786 {coprab 5787 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 105 ax-ia2 106 ax-ia3 107 ax-io 699 ax-5 1424 ax-7 1425 ax-gen 1426 ax-ie1 1470 ax-ie2 1471 ax-8 1481 ax-10 1482 ax-11 1483 ax-i12 1484 ax-bndl 1486 ax-4 1487 ax-17 1503 ax-i9 1507 ax-ial 1511 ax-i5r 1512 ax-14 2115 ax-ext 2123 ax-sep 4056 ax-pow 4108 ax-pr 4142 This theorem depends on definitions: df-bi 116 df-3an 965 df-tru 1335 df-nf 1438 df-sb 1732 df-eu 1993 df-mo 1994 df-clab 2128 df-cleq 2134 df-clel 2137 df-nfc 2272 df-ral 2423 df-rex 2424 df-v 2693 df-sbc 2916 df-un 3082 df-in 3084 df-ss 3091 df-pw 3519 df-sn 3540 df-pr 3541 df-op 3543 df-uni 3747 df-br 3940 df-opab 4000 df-id 4226 df-xp 4557 df-rel 4558 df-cnv 4559 df-co 4560 df-dm 4561 df-iota 5100 df-fun 5137 df-fn 5138 df-fv 5143 df-ov 5789 df-oprab 5790 This theorem is referenced by: (None)
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https://ch.mathworks.com/matlabcentral/cody/problems/42317-de-primed/solutions/1241874 | 1,606,972,188,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141718314.68/warc/CC-MAIN-20201203031111-20201203061111-00483.warc.gz | 191,037,357 | 17,688 | Cody
# Problem 42317. De-primed
Solution 1241874
Submitted on 29 Jul 2017 by Chris Cleveland
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### Test Suite
Test Status Code Input and Output
1 Pass
M = 1:10; M_corr = [2,4,6,4,10,6,14,8,9,10]; assert(isequal(de_primed(M),M_corr))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
2 Pass
M = 20:3:50; M_corr = [20,46,26,58,32,35,38,82,44,94,50]; assert(isequal(de_primed(M),M_corr))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
3 Pass
M = 2:2:100; M_corr = [4 M(2:end)]; assert(isequal(de_primed(M),M_corr))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
4 Pass
M = 10:10:100; M_corr = M; assert(isequal(de_primed(M),M_corr))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
5 Pass
M = 3:3:100; M_corr = [6,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99]; assert(isequal(de_primed(M),M_corr))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
6 Pass
M = eye(4); M_corr = 2*M; assert(isequal(de_primed(M),M_corr))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
7 Pass
M = magic(6); M_corr = [35, 2, 6,26,38,24; 6,32,14,21,46,25; 62, 9, 4,22,27,20; 8,28,33,34,10,15; 30,10,34,12,14,16; 4,36,58,26,18,22]; assert(isequal(de_primed(M),M_corr))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
8 Pass
ind = randi(4); switch ind case 1 M = 1:10; M_corr = [2,4,6,4,10,6,14,8,9,10]; case 2 M = eye(4); M_corr = 2*M; case 3 M = 10:10:100; M_corr = M; case 4 M = magic(6); M_corr = [35, 2, 6,26,38,24; 6,32,14,21,46,25; 62, 9, 4,22,27,20; 8,28,33,34,10,15; 30,10,34,12,14,16; 4,36,58,26,18,22]; end assert(isequal(de_primed(M),M_corr))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
9 Pass
ind = randi(4); switch ind case 1 M = 3:3:100; M_corr = [6,6,9,12,15,18,21,24,27,30,33,36,39,42,45,48,51,54,57,60,63,66,69,72,75,78,81,84,87,90,93,96,99]; case 2 M = 1:10; M_corr = [2,4,6,4,10,6,14,8,9,10]; case 3 M = eye(4); M_corr = 2*M; case 4 M = 20:3:50; M_corr = [20,46,26,58,32,35,38,82,44,94,50]; end assert(isequal(de_primed(M),M_corr))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
10 Pass
ind = randi(4); switch ind case 1 M = 20:3:50; M_corr = [20,46,26,58,32,35,38,82,44,94,50]; case 2 M = 10:10:100; M_corr = M; case 3 M = 2:2:100; M_corr = [4 M(2:end)]; case 4 M = 1:10; M_corr = [2,4,6,4,10,6,14,8,9,10]; end assert(isequal(de_primed(M),M_corr))
[Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.] [Warning: Function assert has the same name as a MATLAB builtin. We suggest you rename the function to avoid a potential name conflict.]
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Start Hunting! | 1,557 | 4,920 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.359375 | 3 | CC-MAIN-2020-50 | latest | en | 0.833842 |
https://www.solutioninn.com/study-help/logic-and-computer-design-fundamentals/in-the-srm-and-slm-instructions-both-the-operand-rsa-1226769 | 1,713,966,268,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296819273.90/warc/CC-MAIN-20240424112049-20240424142049-00054.warc.gz | 899,816,640 | 17,077 | # In the SRM and SLM instructions, both the operand R[SA] and the shift amount ield OP are
## Question:
In the SRM and SLM instructions, both the operand R[SA] and the shift amount ield OP are checked to see if either is 0 before the shifts begin.
(a) Redraw the state machine diagram for these operations with these checks removed.
(b) Use the original diagram and the new diagram to compare the number of clock cycles required for values of OP equal to 0 through 7. Assume that the probability of each OP value for 1 through 6 is 1/8, for 0 is 1/4, and for 7 is 0. Assume that the likelihood of a 0 operand is 1/8. Perform calculations to determine the best implementation (with checks or without checks)
based on the given probability information and comparative number of clock cycles for the two implementations. Provide a convincing argument for your selected answer.
Fantastic news! We've Found the answer you've been seeking! | 224 | 938 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-18 | latest | en | 0.882297 |
https://trustconverter.com/en/base-number-conversion/base-5.html | 1,708,515,660,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473472.21/warc/CC-MAIN-20240221102433-20240221132433-00861.warc.gz | 621,189,080 | 6,442 | base 5 to
decimalbase 5
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510
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813
914
decimalbase 5
1020
1121
1222
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1424
1530
1631
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1934
decimalbase 5
2040
2141
2242
2343
2444
25100
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decimalbase 5
30110
31111
32112
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39124
### Base 5
base 5 is a positional numeral system with five as its base. It uses 5 different digits for representing numbers. The digits for base 5 could be 0, 1, 2, 3, and 4. | 189 | 443 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2024-10 | latest | en | 0.305241 |
http://www.coursehero.com/file/6901617/ACC-505-Case-Study-Springfield-Express/ | 1,411,366,383,000,000,000 | text/html | crawl-data/CC-MAIN-2014-41/segments/1410657136896.39/warc/CC-MAIN-20140914011216-00036-ip-10-234-18-248.ec2.internal.warc.gz | 431,395,874 | 16,104 | This preview has intentionally blurred parts. Sign up to view the full document
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Unformatted Document Excerpt
Springfield Express is a luxury passenger carrier in Texas. All seats are first class, and the following data are available: Number of seats per passenger train car 90 Average load factor (percentage of seats filled)70% Average full passenger fare \$160 Average variable cost per passenger \$70 Fixed operating cost per month \$3,150,000 What is the break-even point in passengers and revenues per month? Breakeven Point in Passengers = Fixed Expenses Contribution Margin Breakeven Point in Passengers = \$3,150,000 (\$160 \$70) Breakeven Point in Passengers = 35,000 Breakeven Point in Revenue = Breakeven Point in Passengers * Average Full Passenger Fare Breakeven Point in Revenue = 35,000 Passengers * \$160 Breakeven Point in Revenue = \$5,600,000 What is the break-even point in number of passenger train cars per month? Contribution Margin = (90 * 70%) * 90 = 5670 Breakeven Point in Passenger Cars = Fixed Expenses Contribution Margin Breakeven Point in Passenger Cars = \$3,150,000 5,670 Breakeven Point in Passenger Cars = 556 passenger train cars If Springfield Express raises its average passenger fare to \$ 190, it is estimated that the average load factor will decrease to 60 percent. What will be the monthly break-even point in number of passenger cars? New Contribution Margin = (Avg Passenger Fare Avg VCPer Passenger) * Avg Load Factor * Number of Seats New Contribution Margin = (\$190 - \$70) * 60% * 90 New Contribution Margin = \$120 *60% * 90 New Contribution Margin = 6,480 Breakeven Point in Passenger Cars = Fixed Expenses Contribution Margin Breakeven Point in Passenger Cars = \$3,150,000 6,480 Breakeven Point in Passenger Cars = 486 passenger train cars (Refer to original data.) Fuel cost is a significant variable cost to any railway. If crude oil increases by \$ 20(Refer to original data.... View Full Document
End of Preview | 455 | 1,993 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2014-41 | latest | en | 0.813887 |
daariss.wordpress.com | 1,547,790,876,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659890.6/warc/CC-MAIN-20190118045835-20190118071835-00212.warc.gz | 470,212,013 | 32,595 | # Our Multiplication Strategies
We have started to apply our understandings of place value and multiplication and extending our knowledge to multiplying two by two digit and three by two digit numbers.
In our classroom, there is no “right” or “wrong” strategy, it is about what works for you. Whichever strategy a child feels most comfortable with and understands not just the application of but the why of, is the best.
We have shared and learned three different multiplication strategies and the amazing Jessie Krefting has created mini-videos demonstrating some of them. Students do not have to demonstrate mastery in all strategies, but must find one they understand and can apply when required.
1. Standard Algorithm
2. Caroll Diagram
3. Partial Product and Why Is Math Different Now by Dr. Raj Shah
Students have now started an inquiry math project called Resort Report.
All mathematical applications will be completed on paper so students can demonstrate their understandings, then uploaded to their document for a complete view of their learning.
# Classroom Learning – Term Two Update
Hi everyone,
I hope you all have had a wonderful holiday break and a great first week back to school! We eased back into our routine last week and have begun some new projects. Below is a brief overview of the learning focus for in our room for this term:
Language Arts
We are finishing the book Out Of My Mind for our read aloud and students will continue making predictions, connections, inferences and asking questions for our reading strategies. They will post their reflections on their blogs and this term we will have a deeper focus on conventions in their writing. Emphasis and assessment will be placed on capitals and punctuation use in all writing.
We will also be focusing on the Organization trait in writing and students will be learning how to gather their thoughts and compose them in a fluent way. We will start a group read aloud on the book Wendell The World’s Worst Wizard and students will have the opportunity to write their own stories with detailed beginning, middle and end while also connecting predictions, inferences and questions with other classes also reading this book.
Students will also start their own book clubs and literature circles this term. Books will be chosen by student groups so as to focus on their interest but also their instructional levels. This may involve some at home reading if a student is not able to read in time allotted in class time. They will meet daily with groups and discuss the book, its contents, their reflections and predictions. Individual at home reading is still to continue whenever possible and is to be recorded on Good Reads or in their reading duotang – whichever they have been assigned. Students are encouraged to continue to review their books, provide recommendations and search for additional books they want to read on Good Reads as well as set their 2015 Reading Challenge goals.
We had a beautiful Reading Tree drawn and painted by the incredibly talented artist Aiden. which we cut out and placed in our room. Students have begun to fill it up with all of the books they have read and rated. Our goal is to have it overflowing by June with all of the books we have read!
Mathematics
We are working on our multiplication strategies with integrated division strategies so students can make the deeper connections between the two with a focus on fact families and deep meaning and understanding of multiplication and division. Students are also working in groups on Tuesdays and Wednesday with Mrs. Krefting‘s class to strengthen their mathematical understandings. We also are continuing with our Problem Of The Week every Fridays and students will reflect on their processes on their blogs.
This term we have also started participating in the Math Photo A Day Challenge with other classrooms around the world. Students are provided with mathematical prompts daily and they are to find and represent their understandings through a photo. We have begun to tweet our photos on Twitter using the hashtag #MathPhotoADay and I will curate all of their photos monthly and post them to this blog for you!
We are also learning how to use charts and graphs by integrating our understandings with an upcoming Science unit of Weather. Students will be graphing weather trends globally.
Science
We began our Classroom Chemistry unit and students have started learning about the States of Matter : Solid, Liquid and Gas. We have watched a Bill Nye video and they have implemented their jot note skills on the states of matter. Students will begin experiments this week with key learning features: mixtures, crystallization, properties of matter and chemical reactions.
Social Studies
Students have worked so incredibly hard on their family history presentations and we are all so honoured to learn about one another’s ancestries and family backgrounds. They began their presentations to our class this week while at the same time received feedback from both myself and two peers. They have been using an assessment tool and interactive feedback document in their Google Docs which allows them to leave and receive peer and teacher feedback. As a student presented, two peers were leaving them feedback and everyone else was practicing their jot note taking and picking out the most important features of their peer’s presentation. I have never been more proud of the hard work and dedication they are demonstrating!
Presenting, providing feedback and taking jot notes!
Working hard on our presentations and our feedback/assessment documents.
Providing feedback and taking jot notes, working on our 1000 piece Map of Canada puzzle and presenting!
We will be meeting with Mr. Kolody and Mrs. Krefting’s class this week and re-presenting our family histories in small groups and making connections between the histories of one another.
This term we will also be focusing heavily on the Regions of Canada. There are six regions: Atlantic, Arctic, Plains, Canadian Shield, Great Lakes St. Lawrence and The Cordillera. We will be diving deep into the Arctic and the Great Lakes St. Lawrence regions and then collaborating with our peers in Mr. Kolody and Mrs. Krefting’s class to make the connections between the other four regions. Students will research and participate in a variety of activities to deepen their understanding of each region. They will then write a persuasive argument for one of the regions and present to the LC 5 groups in a Socratic Circle.
Students who are participating in the Minecraft region creation can continue and share their building as they go.
I will continue to update you on our learning adventures as we continue!
Miss D. Ariss
# Classroom Learning – November 24 – November 28, 2015
November has flown by in our room and we are looking forward to bringing together all of the incredible projects and learning we have been doing this term.
The following is an update on the activities happening in LC5B!
Language Arts
We have started a group read- aloud on the book “Out Of My Mind“. Its an incredibly powerful story of an 11 year old girl diagnosed with Cerebral Palsy and the social and academic challenges she faces. The students have made some deep connections with the main character and are learning how to express empathy and understanding. I am reading this book to them and together we are discussing our thoughts and feelings. The students have also started to blog their own questions and understandings.
We are continuing with our daily reading and ensuring every student has access to reading materials reflective of their interests and ability. Please do let me know if students are not bringing any material home for additional reading time when possible.
Mathematics
We are shifting our main focus to multiplication now, however place value and estimation will still be a continued focus year-round to ensure understanding. The students are also meeting in three smaller groups for two periods a week with Mrs. Krefting’s class to work on combined skills and we continue with our problem of the week every Friday.
This week’s problem was called Grasshopper Jump Fest and I was so proud to see everyone just dive into solving it in their own ways. Students have now built the confidence to attempt problems without fear or hesitation from the work we began at the start of the year. We are working now on how to explain our thinking and how we are actually solving problems. This involves a strong understanding of one’s own strategies and how and why they applied those.
We all have our own strategy for solving the problem!
Students showed a lot of pride in demonstrating their understandings.
Focused and enjoying the level of difficulty of the problem!
Chose to use Google Draw to work through her problem!
A few of the students were so proud of their process and in demonstrating their understandings that they asked me to video them solving the problem. As you can see below, some succeeded and the ones that didn’t during this taped version, continued on despite that fact and then refined their process with the feedback from myself and their peers. At the end of our lesson, they had all achieved success and began their reflection blogs.
Science
We are wrapping up our units on Electricity and Magnetism. The students are working on their final projects which is to build a mechanism that features a working circuit and an element of magnetism. They have all done their research and will begin to plan their experiments this week. Our next unit is weather which will be a year-long focus as weather is happening every day around us. We have started watching and interpreting the weather radars on the weather network as well as tracking weather systems heading our way and globally. The students were fascinated by the extreme snow conditions in Buffalo, New York last week and began to question how being surrounded by lakes causes a Lake Effect. They also started to make connections between that and our location in Alberta versus British Columbia.
Social Studies
The students are now working on putting together their presentations with the information collected from you about their histories and heritage. They refined their questions and understandings last week and started exploring different presentation options. They were provided with a few technology tools as well as poster and paper options and were asked to choose which style worked best for them to represent their families’ story.
Art
We love art in LC5B! Its never an easy task but it allows for students to demonstrate their own creativity. We worked on the use of chalk pastels and the aspect of defining space. Students are often asked to fill in every piece of a project and not to leave any white spaces, however with our chalk pastel pieces, that white space helps to define texture. Students were pushed and directed to actually colour outside of the lines and to make their images rough and not refined.
These are their scruffy Woozles:
Taking ownership and hanging their art pieces!
Working together to display their work!
Nadine’s combination of colour and texture wowed us all!
Really looking forward to this week!
Miss D. Ariss
# Classroom Learning – November 3 – November 7, 2015
We have had quite the event-filled weeks leading up to our current Fall Break. I sincerely hope everyone is resting and gearing up for the next few weeks of learning!
Here is a brief update on the learning activities happening in our classroom:
Mathematics
We continue to review and reinforce the three estimation strategies of FrontEnd, Compatible and Compensation. These are new concepts to students as most are familiar with the use of compatible which allows them to estimate or round to the nearest 10, 100, 1000, 10,000 or 100,000 but there are still struggles with rounding beyond 100 and we will be continuing to review and apply. Students will need to know when and how to apply these strategies when adding and subtracting whole numbers up to 1,000,000. They have demonstrated growth and understanding with addition, however subtraction across multiple place values is an area of focus.
Students are also completing their final place value and estimation project which they will have on padlet.com. They will be posting these to their blogs very soon. Here is a video made by Nate and Ashley describing this project with photos of Ian’s completed project. It also features a student from Mrs. Krefting’s grade 5 classroom sharing about the Language Arts learning we are doing as well:
Language Arts
We have been focusing heavily on the writing trait of Ideas. Students observed a random object within their environment and listed the details. Then as a group we wrote a descriptive paragraph imagining ourselves as that object. Students then showcased their creativity as they assumed the role of their object and described themselves from its perspective. They posted these on their blogs and were so happy to see your guesses in the comments section. Some were very tricky but also quite descriptive.
We also read the book called Nothing Ever Happens On 90th Street and had a group discussion about whether things ever happen at our school. We discussed how we can observe things in our everyday environment so that it may help to spark ideas for our writing. Students then each chose a staff member in our school to visit and observe as they taught their class or worked in our front office. As they observed, students had to write what this person does and says. If this person required specific tools and what their working environment was like. They also had to imagine what superpower they would gift this individual and why and how it would affect them.
Students also had the opportunity to watch a short video called Ideas Are Scary and write a reflective piece on their observation and comprehension of its basic idea. I am trying to push them out of their comfort zones this year so that they can bring out their creativity and showcase their understandings in deeper more meaningful ways. These blog posts were truly inspiring to read.
Our Global Read Aloud is coming to an end as well. We are almost done reading The Fourteenth Goldfish together. The discussions and questions that have come up from this book have truly allowed for us to have very informative discussions surrounding Science and famous scientists. We will be sad to finish this book.
Science
Electricity and Magnetism is really bringing out the student’s love of experimentation. We have discussed quite a bit of information so that students have a strong foundation of content to apply to their experiments. They have learned about conductors and insulators, cells, open and closed circuits, switches, symbols and components. They also are learning how to use the Scientific Method when conducting experiments ensuring they have documented the following: Question, Hypothesis, Materials, Procedure, Observations and Conclusion.
Their very first experiment was to build a circuit and conduct electricity through the use of acidic vinegar, copper wire and metal nails to light a single LED bulb. I didn’t provide students with the exact way to conduct this as I wanted to observe their initial understandings from what we had learned previously about circuits. The objective in this first experiment was to see whether students have fully grasped the concept of an electric circuit, how to apply the Scientific Method when experimenting and what they do when and if they should fail. We have had numerous conversations about scientists in our classroom and how they are continuous learners who fail, but learn from each situation to make their following experiments better.
Not a single group was able to get their bulb to light, which I expected, however every single group continued to change the variables within the experiment and learned from every situation they had attempted. Only one group from Mrs. Krefting’s class got their bulb to light and they came in and shared their findings with us. Students documented their experiment and shared with me.
Here are a few photos and videos of their experiment:
Working together on creating a functioning circuit.
Reconfiguring their circuits numerous times.
Our next step is to create fully functioning circuits and applying the skills learned.
Social Studies
Students have brought back their information from the interviews they conducted with you and their extended families. This was the very first step in our inquiry project into finding out more about our histories and backgrounds. Every student has brought varying amounts of information ranging from a few short answers, in-depth multi-person interviews to detailed family trees. Our next step is to review their content and provide one another with feedback as to how they can get deeper answers so that their information starts to build a complete story instead of random short facts. Once this feedback has been provided, please expect that students will come back to you with more detailed questions in order to improve on their first attempt. Once completed, they will be required to organize and assess their information in order to bring it together into a presentation to share with our other LC5 communities. Students will be provided with various methods of presentation and will have the opportunity to choose which method suits them best.
We have also reviewed our country and its location in the world. We also looked up our family names to find out what they might connect us to. They are very intrigued by the fact that their last names are connected to so many things in the world and have an extensive amount of questions as to who their relatives might have been. These would be wonderful discussion questions to have and learn together about at home.
Stay tuned for more updates this break!
Miss D. Ariss
# How will you use \$5,000,000?
Working on their place value project.
Our focus this year is on real-world applications of the mathematical foundations we build in our classroom. We have been working on understanding place value concepts to 1,000,000 and three strategies (front end, comparable and compensation) for estimation during Math. We’ve also been thinking of how and where we would need to understand and use these concepts in our lives.
The students were then presented with the following problem:
“You have inherited \$5,000,000 from a long lost relative in your ancestor’s country. The problem is that you must use this money on purchasing 3 homes in Alberta. You can use http://www.realtor.ca. One house must be over \$1,000,000 and the other two can be of your choosing, however you must share your reasoning as to why you have chosen these homes. Any money left over will be yours to keep, so choose and estimate your budget wisely. How will you use your money?”
They were also presented with website called Padlet, which is a virtual collaborative board, open to representing learning and projects in a variety of ways.
Students had to organize their three homes and their features. They were to also calculate the estimated value and cost of each home as well as the property taxes. Then calculate their estimated leftover amounts after purchases were processed. Once their calculations and reasonings were completed, they will draft a cheque for their homes using written and standard form for the amount they would pay home owner. Their learning and understandings, whether visual, oral or written along with their calculations would all be posted to their padlet wall for sharing.
Students were ecstatic to start exploring and were even more invested when they saw the connections between the google map they were using in Social Studies and the one on realtor.ca. They started out small and searched for homes within their city and soon afterwards started to branch out.
This is where things started to get exciting because they began to see just how much money a home can cost and some began to have questions.
“Do I only have to choose 3 homes?”
“What if I want two or more homes over \$1,000,000?”
I thought about those questions and immediately questioned my reasonings behind creating the problem. I looked at them and said, “No, you have free choice. It is your \$5,000,000 and so use it at your discretion so long as you can demonstrate your understandings.”
As soon as they realized they had ownership of the project and their money, their interest began to shine through which showed me that they were truly invested in this project and in applying the classroom concepts to the real world. Questions then became these (and to which I said yes):
“Can I purchase multiple rental homes and rent them out for more money?”
“I love art. Do I have to buy a house? Can I l buy art gallery spaces?”
They were making this project their own! Soon all you could hear in our classroom were:
• Students sharing the costs of their homes which meant reading the larger numbers and vocalizing them to one another
• Students using descriptive language to justify which properties they have chosen
• Students learning about different parts of our province and questioning housing costs relative to locations
We even held discussions about why some properties were listed at \$1 and what bidding wars were.
This project has helped to begin the conversations about how math is a part of our everyday lives. That math is not just something you need to “do” in class but that there is a real need for the skills you are learning, and where, why and how to apply them.
# Classroom Learning – October 6 – October 9, 2014
Here is a quick review of the week that passed and what to expect for this upcoming week!
Mathematics
We are continuing to work on problem solving and we started the week with this visual map of a Ghost in the Mansion. I posted the visual on the SmartBoard and students immediately tried to figure out how to manoeuvre from one area to another while also discussing how and why they would choose each route.
Working in groups to find strategies to move through the mansion.
As students worked together to find the best routes, I decided to make it trickier. I wanted them to apply their thinking beyond the problem written on a piece of paper. I told them I was providing each group with a roll of tape and wanted to see if they could re-create the map on a larger scale and then manoeuvre themselves through it.
Now this brought up many more problems for them beyond the mathematical application as they needed to work together effectively to produce a common group outcome. We decided to move out into our pod and use the carpet space provided.
Starting to figure out how to apply skills on a larger scale.
Getting past the struggles to create their models.
Working in a group proved to be a huge challenge for most as each had their own idea and vision for how their models would look. Some groups worked much more effectively and were able to listen to group member’s ideas and ensured that all were involved. Others hit major communication roadblocks which hindered their progress. This activity transformed itself from a math problem to a social and group work problem and some very valuable lessons were learned. I was very proud though of each group for being honest about what their struggles were and in trying to apply strategies to solve. In the end, some groups decided to part ways and others remained in tact. Students also wrote individual reflections for me regarding what challenges they faced during this activity and how it affected their ability to move forward.
Now that the blueprint is created…how do we make our way through it?
I was really amazed to see them work their way through their larger scale models. It was much harder for some to visualize the map while also moving themselves through it. Just a simple transformation of a problem, allowed them to view it much differently and made for a different set of strategies.
This week, we will review this problem again and reflect on our learnings. The students have also completed a mini review of place value and all have shown their knowledge and so we will be moving forward with estimation strategies. We will be beginning a project based on real-life applications of place value and estimation focused on purchasing property in Alberta.
Social Studies
We have been learning about our heritage and who we are. Our theme for this year is ” Who Do You Think You Are?” as we also focus on Canada as a country and its foundations. We had a lot of discussions about where we think our ancestors came from and how they may have come to Canada. One of our starting projects was the one sent home with the students this weekend. To prepare them for asking deeper questions that transcend beyond the “what is your favourite food?” and more along the lines of “where is my family from and how and why did they come to Canada?”, students were paired up to interview one another. One student was the interviewer, another was the interviewee and a third was the question collector.
Interviewing one another!
We brainstormed together just how we could transform the what, where and when questions into deeper why and how questions. Then we paired up with student’s from Mrs. Krefting’s grade 5 classroom to watch an episode of the show “Who Do You Think You Are?” where actor Jim Parsons from The Big Bang Theory discovers his heritage by asking deeper questions to lead him on his quest for finding the answers.
Collaboratively applying our interviewing skills.
This week we will review their findings from their home interviews, discuss deeper questioning and fill in gaps as needed for interviewing at home before beginning to piece together the information.
Science
We are learning about the connections between electricity, magnetism and static. Students each received a balloon, a variety of different materials and provided with an outcome but not with the “how” or “why” to experiment with their findings. Some of the outcomes were:
• Make the balloon stick to the wall without using anything else for help
• Make a cereal pendulum swing/bounce without touching it, blowing on it or shaking the table
• Make the cereal jump/move without touching them, blowing on them or shaking the table
• Turn the tap on so that a small stream of water is running out, make the stream bend without touching it or blowing on it
Students recorded their findings along with how they did it, why they think it worked and any questions they had regarding it. Some of the ways they came up with for moving their materials through static electricity were absolutely incredible.
Video – She decided to crush her cereal and use her balloon as a magnet
Video – Using the cereal to move her balloon
Video – Realizing how she could move the balloon with her leg
Instinctively using balloons to create static electricity!
This week we will be focusing on electricity and circuits.
Language Arts
We are participating in the Global Read Aloud with the book The Fourteenth Goldfish. Students have been making predictions and tweeting their thoughts via our Twitter account. We have been connecting with a variety of classrooms online about the story and what we think is going to happen. Students have also began to utilize their blogs for reflection on the story and a variety of other topics.
They were overjoyed to see their parents leaving them comments last week and interacting with their learning. They immediately wanted to read not only theirs but their peers’ blogs too. They have been reading, reflecting and providing each other with comments and feedback and welcome you to join us!
They will be blogging daily and we look forward to connecting with you!
This was a long review due to parent conferences last week but I sincerely appreciate your time and feedback. Please let me know below in the comment section your thoughts!
Some questions to ponder and discuss with your child and in the comment section below:
• Where do you use place-value and estimation strategies in your daily life?
• Where in your home do you have electric circuits and can you share these with your child?
We look forward to hearing your feedback as a class!
Miss D. Ariss
# Classroom Learning – September 26, 2014
We have jumped right in to building our foundation for life-long learning in LC5B. As our core focus in on inquiry and problem-solving through real-world applications, the students are learning how to ask questions and how to approach problems in every aspect of our curriculum.
To have students thinking about problems and ways to apply different strategies that work for them, we looked at assisting this Rocket Propelled Coyote from one place to another. Students came up with their own individual solutions through actively trying, failing and re-attempting different possible outcomes. I loved watching them push through their frustrations and share with one another the ways in which they solved the coyote’s problem.
Figuring out how to get the coyote from one place to another.
At one point during our session, Mr. McLean joined us and pointed out that he had a different interpretation of the problem. He was attempting to get the Coyote to his exact location, while we were attempting to get him there within one 1km. That moment allowed us to learn that how we read and interpret a problem may be different which would also produce different outcomes.
Our challenge to you….can you help the Coyote and show or tell us how you did it? Please leave us a comment below! | 5,754 | 29,718 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.9375 | 3 | CC-MAIN-2019-04 | latest | en | 0.948092 |
https://www-en.fisica.uniroma2.it/teaching-modules/theory-of-solids-and-molecular-models/ | 1,721,792,742,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518154.91/warc/CC-MAIN-20240724014956-20240724044956-00200.warc.gz | 555,406,030 | 7,163 | ## Theory of Solids and Molecular Models
course ID
Lecturer
CFU
8
Length
14 Weeks
Semester DD
First
##### Course details
The Born-Oppenheimer approximation
The Hellmann - Feynman theorem , Epstein theorem.
Band theory in solids Bloch theorem , boundary conditions variational method , tight-binding method and its applications – Ortogonalized Plane Waves, pseudopotentials .
Ab-initio methods : Hartree and Hartree Fock equation , Koopmans theorem , potential for gas exchange electronic homogeneous Fourier Transform Coulomb potential of the homogeneous electron gas with the Hartree Fock. Approximation of Slater, Thomas Fermi approximation. functional derivatives
Density Functional Theory . Theorem of Hohenberg and Kokn , Kohn and Sham equations .
The Local Density Approximation . The problem of the gap in DFT
Examples of applications of DFT
Optical properties
Complex refraction index. The absorption coefficient.
The reflectivity . The dielectric function . Kramers Kroning relations and sum rules
Fermi's golden rule : Calculation of the dielectric function in the dipole approximation
Examples of dielectric function for metals, semiconductors, insulators . Joint density of states ( JDOS ) and it behaviour near the critical points.
Linear response theory and TDDFT .
Boltzman equation for electric and thermal transport
Classical and ab-initio molecular dynamics
Excitonic effects : model hydrogen- Mott – Wannier.
Ab-initio excited state theories Classical Green's functions . Formalism of second quantization . Quantum propagator of a single electron / hole and its representation of Lehmann and relationship with electronic excitations . Dyson equation . Self- energy concept . Quasi-particle Equation. GW method . Bethe- Salpeter equation for the calculation of excitonic effects in the optical response .
Practical lessons at computer of DFT , TDDFT , GW and BSE
which include an introduction to the main commands in linux environment
Co-teaching: Prof.ssa Palummo Maurizia
##### Objectives
LEARNING OUTCOMES:
The course is aimed at completing basic training in the field of quantum physics applied to the study of microscopic and macroscopic properties of materials.
The goal of the course is to provide the main knowledge on theoretical / computational methods for the study of the structural, electronic and optical properties of materials.
The main educational objectives are the understanding of quantum-mechanical semi-empirical and first-principles methods, such as the Density Functional Theory (DFT), the time-dependent Functional density theory and the Green Function theory.
Another objective is the learning and autonomous use of one of the main DFT (quantum-express) calculation codes currently in use in the field of research in material science through the performance of practical exercises by the student.
KNOWLEDGE AND UNDERSTANDING:
The course aims to provide the student with the basic tools needed to understand the structural and opto-electronic properties of materials in terms of a microscopic quantum-mechanical description.
The lessons focus on the mathematical derivation and physical interpretation of the main theoretical investigation tools for the study of the structural, electronic and spectroscopic properties of materials.
Applications relating to materials of current interest in the field of material science research are illustrated during lectures and computer-based simulations in order to broaden the student's knowledge about the state of the art in this field.
APPLYING KNOWLEDGE AND UNDERSTANDING:
The course aims to provide mathematical-physical tools that allow students to understand scientific manuscripts dedicated to the study of materials and to analyze, through their knowledge, various experimental physical observables of interest in materials science.
The student must also be able to identify and understand the theoretical / computational method suitable for the characterization of the chemical-physical properties of the material of interest and tp be able to understand, analysis, discussions and data derioved by these methods.
The student will also be able to tackle new scientific problems and to read scientific texts and articles in English on topics related to the study of electronic structural and optical properties of materials.
MAKING JUDGEMENTS:
Students are required to use the acquired knowledge in a critical manner, specifically to study the structural, electronic and optical properties of materials in order to evaluate their characteristics for an appropriate use in the field of materials science.
COMMUNICATION SKILLS:
Particular attention is paid to the ability to use the knowledge acquired during the lessons appropriately and in a conceptually coherent and rigorous context. The final report related to the computer simulations carried out by the student on a specific material, is foreseen through a seminar-type power-point presentation by the same, and has the purpose of exercising and improving communication skills and transversal skills of the student .
LEARNING SKILLS:
Particular attention is paid to the ability to use the knowledge acquired during the lessons appropriately and in a conceptually coherent and rigorous context.
The final report related to the computer exercise carried out by the student on a specific material, is foreseen through a seminar-type power-point presentation and has the purpose of exercising and improving his/her communication skills and transversal skills . | 1,049 | 5,526 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2024-30 | latest | en | 0.796185 |
https://coding-gym.org/challenges/find-kth-zero/ | 1,726,494,103,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.32/warc/CC-MAIN-20240916112213-20240916142213-00703.warc.gz | 146,086,948 | 9,414 | # Find Kth Zero
See the original problem on HackerRank.
Consider an array of integers A = [a1, a2, …, aN]. We can perform two types of queries on A:
• 1 k: For query type 1, find and print an integer denoting the index of the kth zero in array A on a new line; if no such index exists (i.e., there is no zero), print NO instead.
• p x: For query type 2, replace the element positioned at index p with integer x (i.e., set A[p] = x).
Given A’s initial state and queries, perform each query in order. For each query of type 1, print the query’s answer on a new line.
### Input Format
The first line contains two space-separated integers describing the respective values of n (the size of the array) and m (the number of queries). he second line contains space-separated integers describing the respective elements of array A (e.g. a0, a1, …, aN).
Each of the m subsequent lines describes a query in one of the following forms:
• 1 k, where 1 is the query type and k denotes the kth zero you must locate an index for in the array.
• 2 p x, where 2 is the query type, p is the index (position) you must update, and x denotes the new value to store at A[p].
### Constraints
1 2 3 0<= A[i], x<= 10'000 1<= n,k,m <= 10'000 0<= p < n
### Output Format
For each query of type 1, print a single integer on a new line denoting the index where the kth zero is located in A; if no such index exists (i.e., there is no zero), print NO instead.
## Solutions
### Naïve solution
The naïve solution is to create the array just as it defined and do a linear search for each query.
Although update is fast ($$O(1)$$), the search is slow, because we need to search linearly the kth zero ($$O(N)$$).
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 #include #include int main() { std::string line; int n, m; std::cin >> n; std::cin >> m; std::vector arr; while (n--) { int tmp; std::cin >> tmp; arr.push_back(tmp); } while (m--) { int queryType; std::cin >> queryType; if (queryType == 1) { int k; std::cin >> k; int found = 0; int i = 0; for (; i != arr.size(); ++i) { found += arr[i] ? 0 : 1; if (found == k) { break; } } if (found == k) { std::cout << i; } else { std::cout << "NO"; } std::cout << "\n"; } else if (queryType == 2) { int p, x; std::cin >> p; std::cin >> x; arr[p] = x; } } }
### Segment tree
In this Problem, we have two types of queries that are to be performed on the array. One is called as Kth Zero over a range and other is a Point update Query. For this class of problems, the suitable data structure is a Segment Tree.
Thus, the quickest solution uses such a data structure (credits to HackerRank):
Build the tree. Node value is the count of zeroes in the range.
Find 2nd zero
Change value at position 1 to 5
#### Segment tree (C)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 #include void buildTree(int tree[], int array[], int index, int low, int high) { if (low == high) { if (array[low] == 0) { tree[index] = 1; } else { tree[index] = 0; } } else { int mid = (low + high) / 2; buildTree(tree, array, index * 2, low, mid); buildTree(tree, array, index * 2 + 1, mid + 1, high); tree[index] = tree[index * 2] + tree[index * 2 + 1]; } } void updateTree(int tree[], int index, int low, int high, int pos, int new_val) { if (low == high) { if (new_val == 0) { tree[index] = 1; } else { tree[index] = 0; } } else { int mid = (low + high) / 2; if (pos <= mid) { updateTree(tree, index * 2, low, mid, pos, new_val); } else { updateTree(tree, index * 2 + 1, mid + 1, high, pos, new_val); } tree[index] = tree[index * 2] + tree[index * 2 + 1]; } } int find_kth(int tree[], int index, int low, int high, int k) { if (tree[index] < k) { return -1; } if (low == high) { return low; } int mid = (low + high) / 2; if (tree[index * 2] >= k) { return find_kth(tree, index * 2, low, mid, k); } else { return find_kth(tree, index * 2 + 1, mid + 1, high, k - tree[index * 2]); } } int main() { int n, m; scanf("%d%d", &n, &m); int arr[n], tree[4 * n + 1]; for (int i = 0; i < n; i++) { scanf("%d", &arr[i]); } buildTree(tree, arr, 1, 0, n - 1); for (int i = 0; i < m; i++) { int type, k, pos, val; scanf("%d", &type); if (type == 1) { scanf("%d", &k); int ans = find_kth(tree, 1, 0, n - 1, k); if (ans == -1) { printf("NO\n"); } else { printf("%d\n", ans); } } else { scanf("%d%d", &pos, &val); if (pos < n) { updateTree(tree, 1, 0, n - 1, pos, val); } } } return 0; }
#### Segment tree (C++)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 #include #include void update_leaf(std::vector& tree, int index, bool isZero) { tree[index] = isZero ? 1 : 0; } void update_node(std::vector& tree, int index) { tree[index] = tree[index * 2] + tree[index * 2 + 1]; } void build_tree(std::vector& tree, const std::vector& arr, int index, int low, int high) { if (low == high) return update_leaf(tree, index, arr[low]); const int mid = (low + high) / 2; build_tree(tree, arr, index * 2, low, mid); build_tree(tree, arr, index * 2 + 1, mid + 1, high); update_node(tree, index); } void update_tree(std::vector& tree, int index, int low, int high, int pos, bool isZero) { if (low == high) return update_leaf(tree, index, isZero); const int mid = (low + high) / 2; const bool goLeft = pos <= mid; const int nextIndex = index * 2 + (goLeft ? 0 : 1); const int nextLow = goLeft ? low : mid + 1; const int nextHigh = goLeft ? mid : high; update_tree(tree, nextIndex, nextLow, nextHigh, pos, isZero); update_node(tree, index); } int find_kth(const std::vector& tree, int index, int low, int high, int k) { /* If in this branch there are less than k zeros, abort searching */ if (tree[index] < k) return -1; /* If in a leaf, we've found it! */ if (low == high) return low; /* Search in subtree */ const int mid = (low + high) / 2; const bool goLeft = tree[index * 2] >= k; const int nextIndex = index * 2 + (goLeft ? 0 : 1); const int nextLow = goLeft ? low : mid + 1; const int nextHigh = goLeft ? mid : high; const int nextK = goLeft ? k : (k - tree[index * 2]); return find_kth(tree, nextIndex, nextLow, nextHigh, nextK); } int main() { int n, m; std::cin >> n >> m; std::vector arr; std::vector tree(4 * n + 1); for (int i = 0; i != n; ++i) { int x; std::cin >> x; arr.push_back(x == 0); } build_tree(tree, arr, 1, 0, n - 1); while (m--) { int type, k, pos, val; std::cin >> type; if (type == 1) { std::cin >> k; int ans = find_kth(tree, 1, 0, n - 1, k); if (ans == -1) { std::cout << "NO\n"; } else { std::cout << ans << "\n"; } } else { std::cin >> pos >> val; update_tree(tree, 1, 0, n - 1, pos, val == 0); } } return 0; }
### Collection of zero’s indexes
However, the space cost is 4-times the problem dimension. A possible solution which allocates at most N elements but is ~4-times slower than the previous one consists in keeping track only of the zeros updates:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 #include #include #include #include #include using namespace std; int main() { int n, m, elem; cin >> n >> m; deque zeros; for (auto i=0; i> elem; if (elem == 0) zeros.push_back(i); } int cmd, a1, a2; while (m--) { cin >> cmd; switch(cmd) { case 1: cin >> a1; --a1; cout << (a1>=zeros.size() ? "NO" : to_string(zeros[a1])) << "\n"; break; case 2: cin >> a1 >> a2; auto it = lower_bound(begin(zeros), end(zeros), a1); if (it != end(zeros) && *it == a1 && a2!=0) { zeros.erase(it); } else if (a2 == 0) { zeros.insert(it, a1); } break; } } }
Note: we used a std::deque that is, basically, a paged array. Replacing it with a std::vector is twice as slower.
Same algorithm but in Rust courtesy of Alessandro and Edoardo:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 fn main() { use std::collections::BTreeSet; use std::io::{self, BufRead}; let stdin = io::stdin(); let handle = stdin.lock(); let mut lines = handle .lines() .skip(1) .map(Result::unwrap) .map(|s| s.trim().to_string()); let mut zero_positions: BTreeSet<_> = lines .next() .unwrap() .split(' ') .enumerate() .filter(|&(_, s)| s == "0") .map(|(index, _)| index) .collect(); lines.for_each(|line| { let mut splitted = line.split(' '); match splitted.next().unwrap() { "1" => { let index = splitted .next() .unwrap() .parse::() .unwrap() - 1usize; match zero_positions.iter().nth(index) { Some(pos) => println!("{}", pos), None => println!("NO"), } } "2" => { let index: usize = splitted .next() .unwrap() .parse() .unwrap(); let value: u32 = splitted .next() .unwrap() .parse() .unwrap(); if value == 0 { zero_positions.insert(index); } else { zero_positions.remove(&index); } } _ => unreachable!(), } }) }
We've worked on this challenge in these gyms: modena milan padua turin | 3,305 | 9,240 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2024-38 | latest | en | 0.730222 |
https://www.drivethrurpg.com/product/108869/100-numbers-between-1-and-100-Inclusive?language=de&filters=0_2150_300_0_0 | 1,560,903,507,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998844.16/warc/CC-MAIN-20190618223541-20190619005541-00086.warc.gz | 715,862,693 | 44,625 | Erweiterte Suche >
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# 100 numbers between 1 and 100, Inclusive
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Have you ever found yourself stuck mid-game, searching for a random number between 1 and 100 but unable to come up with one?
Be stuck no more!
The Most Mathematically Correct Lists Presents:
100 numbers between 1 and 100!
Never be at a loss for numbers between 1 and 100 again. This handy reference will undoubtably be useful to both players and game masters as they strive to introduce a more random element into table top gaming.
A few examples of what you'll find within:
43. 43
17. 17
98. 98
And 97 other numbers! The book also features a picture of sliced bread! The greatest thing since... well, since anything else!
This book is system neutral, meaning you can use it with any game system which may require a number located between 1 and 100. Not only that, but should you need a number betwee 1 and 100 even outside of your gaming career - this book will do that!
Amaze your friends and co-workers, impress the boss and be a veritable font of guaranteed mathematically correct numbers between 1 and 100!
This is a Creative Commons licensed product.
Kunden, die dieses Produkt gekauft haben, kauften auch
Rezensionen (24)
Diskussionen (3)
Jeffrey K October 01, 2016 8:28 pm UTC
Which entries are the most popular?
Ben G October 02, 2016 12:41 am UTC
PUBLISHER
14, 67 and 92 seem to be the most popular.
March 18, 2016 11:48 am UTC
I'm left with multiple questions: Are the numbers a random sampling, or do they fall in a Gaussian distribution (or some other distribution)? Is there any place to track errata? If so, will there be a second edition? And if so, will any of the current numbers change? Do I need to worry about the bread growing mold?
Ben G March 18, 2016 12:03 pm UTC
PUBLISHER
It's possible due to a fundamental change in physics as we know it that the numbers may change and thus prompt a second edition. In which case, moldy bread will be the least of your worries.
Kai B May 06, 2016 7:54 pm UTC
Ye cannot change the laws of physics, laws of physics, laws of physics;
ye cannot change the laws of physics, laws of physics, Jim.
jesse M June 11, 2015 12:38 am UTC
Will this book be part of a series, or is it a stand-alone product?
Ben G June 11, 2015 12:51 am UTC
PUBLISHER
If I happen to sit down and create some more numbers, I'll add to it. Right now though I'm negotiating the rights to the HBO series based on my work, which would slow down the process for tabulating future numbers greatly.
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https://www.cryptocoinsinfoclub.com/how-to-calculate-crypto-gains-and-losses/ | 1,652,963,214,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662527626.15/warc/CC-MAIN-20220519105247-20220519135247-00001.warc.gz | 854,155,931 | 60,152 | Wednesday, May 18, 2022
# How To Calculate Crypto Gains And Losses
## Trading Cryptocurrency For Another Type Of Cryptocurrency
Calculating Crypto Gains / Losses
Generally, when you dispose of one type of cryptocurrency to acquire another cryptocurrency, the barter transaction rules apply. You have to convert the value of the cryptocurrency you received into Canadian dollars. This transaction is considered a disposition and you have to report it on your income tax return. Report the resulting gain or loss as either business income or a capital gain .
## How Does Bitcoin Tax Calculators Work
The bitcoin tax calculator shows the capital gains tax on bitcoins depending on the holding period. You must enter the purchase price and the sale price of the bitcoin along with the holding period.
For example, you have bought some bitcoin units in August 2017 for Rs 50,000 and sold them for Rs 1,00,000 in November 2018. The holding period is under three years. The gains are short-term capital gains of Rs 1,00,000 Rs 50,000 = Rs 50,000. It is added to your taxable salary and you are taxed as per your income tax bracket.
Suppose you had purchased some bitcoin units in January 2015 for Rs 1,00,000 and sold them in May 2018 for Rs 5,00,000. The holding period is above three years. The gains are called long-term capital gains and are taxed at 20% with indexation benefit.
The long-term capital gains are:
Original cost of acquisition = Rs 1,00,000CII in the year of purchase = Rs 240CII in the year of sale = Rs 280
You have the indexed cost of acquisition = CII in the year of sale* / CII in the year of purchase.
Financial Year
Indexed cost of acquisition = 280 * / 240 = Rs 1,16,666. The sale price of the asset = Rs 5,00,000
Long-term capital gains = The sale price of the asset The indexed cost of acquisition.
Long-term capital gains = Rs 5,00,000 Rs 1,16,666 = Rs 3,83,333. You have long-term capital gains at 20% = Rs 3,83,333 *0.2 = Rs 76,667.
## Irs Increasing Enforcement Of Cryptocurrency Tax Reporting
The IRS estimates that only a fraction of people buying, selling, and trading cryptocurrencies were properly reporting those transactions on their tax returns, but the agency provided further guidance on how cryptocurrency should be reported and taxed in October 2019 for the first time since 2014.
Beginning in tax year 2020, the IRS also made a change to From 1040 and began including the question: “At any time during 2020, did you receive, sell, send, exchange or otherwise acquire any financial interest in any virtual currency?”
If you check “yes,” the IRS will likely expect to see income from cryptocurrency transactions on your tax return.
Don’t Miss: What Is The Best Cryptocurrency Wallet
## Can I Write Off Lost Or Stolen Cryptocurrency
Occasionally, investors may lose cryptocurrency due to events such as a hack or a lost wallet key.
After the Tax Cuts and Jobs Act of 2017, these types of casualty and theft losses are no longer considered tax deductible.
For more information, check out our guide to reporting lost or stolen cryptocurrency.
## When Is Crypto Taxed As Income In Austria
Some crypto transactions are viewed as additional income – which means theyll be subject to tax at the point you receive the crypto, as opposed to at the point of disposal.
Like above, the rules vary depending on whether you acquired your crypto before the 28th of February 2021 or after. Lets cover both.
Under the old crypto tax rules, mining crypto, bounty rewards and affiliate rewards are considered income. So youll pay Income Tax on mined coins or rewards based on their fair market value in euros on the day you receive them. If you later sell, trade or spend your coins within 1 year – youll pay Income Tax on any profits.
Under the old crypto tax rules, these arent the only activities considered income. Other interest-bearing activities are also considered income. This includes:
• Staking crypto.
• Lending crypto.
• Other interest-bearing activities like yield farming or liquidity mining.
Like with mining – you would pay tax on the fair market value of any coins/tokens on the day you received them. But you dont pay Income Tax, instead youll pay a flat 27.5% tax on these transactions.
Under the new crypto tax reform – all this is simplified. Youll pay the flat 27.5% tax rate at the point you receive any coins/tokens from:
• Mining crypto.
• Shop to earn.
• Crypto faucets.
You May Like: Where To Buy Bitcoin Miner
## What Impact Would An Increase In The Capital Gains Tax Have On Crypto Traders
President Bidens proposal to raise the long-term capital gains tax from 20 to 39.6% only affects those with income over \$1 million. This includes around 0.3% of households, according to White House advisor Brian Deese. As a result, most crypto traders and investors long-term tax rates remain unchanged. The incentive to hold on to cryptocurrency for the long term vanishes for individuals who are affected by the almost doubled tax rate.
This isnt the first time capital gains taxes have increased, considering the Tax Reform Act of 1986 and the American Taxpayer Relief Act of 2012 increased stock sales. However, this pattern suggests that those with large amounts of unrealized profits will sell later due to the lower tax rate on their cryptocurrency earnings. With this proposal, those traders would ultimately pay more in taxes.
While theres still much ambiguity surrounding the proposal, there are ways to use the terms to offset capital gains. This provides lower taxes on bitcoin gains if rates rise.
## The Consequences Of Failing To File Crypto Losses On Tax Returns
Various tax authorities work closely with crypto exchanges to track down crypto tax evaders. However, the information that tax authorities are usually not complete. As a result, it is up to you to keep accurate records. If you do not make updated filings with exact details, the tax agency might end up sending you a letter for capital gains taxes you have not paid.
Due to how often you move crypto, only you have an accurate record backed by evidence regarding your activities in the crypto market. Filing your capital losses helps avoid confusion, which can sometimes take months to resolve and land you in a lot of legal trouble.
Also Check: How To Learn Blockchain From Scratch
## If You Mine Cryptocurrency
Cryptocurrency mining refers to solving cartographic equations to validate and add cryptocurrency transactions to a blockchain. In exchange for this work, miners receive cryptocurrency.
If you earn cryptocurrency by mining it, it’s considered taxable income and will typically be reported on For 1099-NEC at the fair market value of the cryptocurrency on the day you received it just as if it were self-employment income.
## If I Got Paid With Cryptocurrency For Work Done As An Independent Contractor Is It Self
Canadian Crypto Taxes Pt. 2: How To Calculate Your Cryptocurrency Gains or Losses for CRA in Canada
Yes. Generally, self-employment income includes all gross income derived by an individual from any trade or business carried on by the individual as other than an employee. Consequently, the fair market value of cryptocurrency received for services performed as an independent contractor, measured in U.S. dollars as of the date of receipt, constitutes self-employment income and is subject to self-employment tax.
Don’t Miss: How To Exchange Cryptocurrency To Cash
## Fill Out The Proper Tax Forms
Once you have a record of your crypto transactions, youll need to fill out certain tax forms depending on how you used your crypto:
• Form 8949. This form logs every purchase or sale of crypto as an investment. This should include the total number of coins, the date and price you bought, the date and price you sold and your gain or loss for each transaction.
• Schedule D. This form summarizes your total capital gains and capital losses from all investments, including crypto.
• Schedule C. If you received coins from mining, you need to disclose whether you received them as a business or as a hobby. If youre running a crypto mining business, you may owe self-employment taxes if your income exceeded your expenses for the year.
• Schedule 1. If you report your crypto mining as a hobby, youd report this income on Line 8 of Schedule 1. You wont owe self-employment tax, but you become more limited on what you can deduct as an expense.
## Tax On Bitcoin In India
With the rags to riches stories circulating around cryptocurrencies, especially bitcoins, the Centre is seriously contemplating bringing the investors under the tax regime. The Central Board of Direct Taxes has already announced that people who made money out of the bitcoin must declare and pay the relevant tax.
The department is all set to send legal notices to those who refuse to comply with the laws. The government is still inclined towards making bitcoin completely illegal and is awaiting suggestions from the committee appointed for this purpose. Even if it may not be abolished altogether, there will be some kind of a regulator and set tax rate slabs.
Recommended Reading: How To Invest In Bitcoins In Us
## What Happens If I Dont File Crypto Losses On My Taxes
Crypto exchanges report information to the IRS, and crypto investors have received letters and notices from the IRS recommending crypto tax filing and even requests for more taxes paid.
Many of the leading crypto exchanges send Forms 1099 to investors who have had more than \$600 of trades, meaning that the IRS will also receive a report of each trader’s activity. Additionally, even exchanges who do not send 1099s can be compelled to share information with the IRS through a John Doe summons, an investigative tool increasingly used by the Biden administration.
The information the IRS receives from these exchanges is often incomplete, however. For example, if you bought bitcoin on Coinbase, transferred it to a separate foreign exchange, and incurred losses on that other exchange before sending bitcoin back to Coinbase to sell it for USD, then the IRS may only account for that BTC sale.
In this case, the agency doesnt have the information to know that you have an overall capital loss with crypto. By properly calculating your crypto taxes and reporting them to the IRS on Form 8949 and Schedule D, then you will show that you do not have any net capital gains that should be taxed.
## How To Calculate Capital Gains And Losses In A Cryptocurrency
For every trade partial or full you need to know the following details:
• When did you buy coins
• How much did you pay for them
• When you sold coins
• How much did you get for them
• More sophisticated exchanges may have a reporting mechanism to help you gather this information. Otherwise, if you havent kept your own detailed records, you may need to access email, bank receipts, or wallet receipts.
Once you have this information, there are several options for performing mathematical calculations. For example, some investors use a first in, first out methodology in which the first coins you buy are also the first coins you sell. We will not cover all the methods and mathematical calculations here. You can use Google to find out more about capital gains calculation options.
Read Also: How To Buy Bitcoin On Coinbase App
## Losses From Hacks & Thefts
Losses on cryptocurrency investments due to scams, hacks, or other thefts are not treated as casualty losses, but instead, as investment losses.
According to tax code 165 , even though the investment is not linked to any business, your investments have been put in for profit. This is why any loss that has occurred as a result of scams, theft, or fraud is tax losses. But how can you claim such losses when filing taxes?
Crypto losses as a result of hacks and theft can be claimed as \$0 proceeds transactions on Form 8949. This implies that if you paid \$15,000 for 1 ETH and it was taken as a result of an exchange breach, you might claim a loss of \$15,000.
## Keeping Books And Records
If you acquire or dispose of cryptocurrency, you have to keep records of your cryptocurrency transactions. This also applies to businesses that accept cryptocurrency as payment for goods and services.
Cryptocurrency exchanges have different standards for the kinds of records they keep and how long they keep them. If you use cryptocurrency exchanges, we suggest that you export information from these exchanges periodically to avoid losing the information necessary to report your transactions. You are responsible for keeping all required records and supporting documents for at least six years from the end of the last tax year they relate to.
You should maintain the following records on your cryptocurrency transactions:
• the date of the transactions
• the receipts of purchase or transfer of cryptocurrency
• the value of the cryptocurrency in Canadian dollars at the time of the transaction
• the digital wallet records and cryptocurrency addresses
• a description of the transaction and the other party
• the exchange records
• the software costs related to managing your tax affairs.
If you are a miner, also keep the following records:
• receipts for the purchase of cryptocurrency mining hardware
• receipts to support your expenses and other records associated with the mining operation
• the mining pool details and records
Recommended Reading: How To Get Free Cryptocurrency
## Frequently Asked Questions About Cryptocurrency And Capital Gains
Cryptocurrency can be a great currency alternative, but remember that taxes will almost always play a role. Just like your income and the payments you make using U.S. currency, crypto income and purchases are taxed at a similar rate. However, you have several options for using your crypto without incurring taxes as well.
Still have questions? Here are the answers the IRS has provided for frequently asked questions about cryptocurrencies and taxes.
## When Does Capital Gains Tax Apply To Crypto Transactions
Cointracking Tool Helps You Figure Out Bitcoin & Crypto Gains, Losses, and Tax Implications
Not all proceeds from crypto investments are seen as a capital gain or loss. Sometimes proceeds will be subject to Income Tax or even tax free. This is all dependent on where you live, so you should always check your countrys crypto tax rules. In general though, the following rules apply:
Seen as a capital gain/loss – Capital Gains Tax
• Selling crypto for fiat currency.
• Swapping crypto for crypto – depending on where you live.
• Spending crypto to buy things.
• Gifting/donating crypto – depending on where you live.
Seen as income – Income Tax
• Earning crypto.
• Moving crypto between wallets.
Wallet fees
Something to be aware of: If youre moving assets around between your own wallets, there is no profit or loss as youre not buying or disposing of an asset. However, if you paid a transaction fee to carry out a transfer, this fee is often subject to taxes so you may need to factor this into your P& L calculations.
Also Check: How To Check Bitcoin Rate
## Cra Cost Basis Method
The Canadian Revenue Agency says taxpayers must use the adjusted cost basis method when calculating crypto capital gains and losses. The adjusted cost basis method is the cost of an asset plus any fees related to it.
You can either use the fair market value of the asset at the point you acquired it or the FMV of the asset at the end of the year – whichever is lower. For this reason, you need to keep very accurate records of your crypto transactions in Canada.
For investors with multiple assets, you can choose to value your entire inventory for its FMV at the end of the year instead.
You can find more information in our Canada Crypto Tax Guide.
## More From Smart Tax Planning:
Here’s a look at more tax-planning news.
“Be prepared to pay some tax,” said enrolled agent Adam Markowitz, vice president at Howard L Markowitz PA, CPA in Leesburg, Florida.
But calculating your balance can be tricky, he said, particularly if it was a year of heavy trading.
Read Also: How Do I Buy Safemoon Crypto
## Understanding Tax Rate On Crypto Capital Gains
• Key takeaways
• The IRS treats crypto as a digital asset and taxes it much like stocks, bonds, and other capital assets. However, the taxes you owe will depend on how you used your crypto in the previous year. Yes, its a little complicated. Heres what you need to know.
Cryptocurrency investors began to question what impact the presidents proposal to raise the capital gains tax on the rich would have shortly after it launched. The proposal, which intends to raise the long-term capital gains tax rate from 20 to 39.6% for persons earning at least \$1 million in yearly investment income, sparked debate in the cryptocurrency world.
What does this signify for cryptocurrency investors and traders? How can you keep your capital gains taxes to a minimum? Keep reading to find out more!
## Can You Claim A Capital Loss If You Havent Sold Your Crypto
Remember, you need to actually realize your loss for it to count as a capital loss that can be written off on your taxes. To realize a loss, you must incur a taxable eventâin other words, you need to actually dispose of your crypto to realize the loss.
Examples of disposals include the following:
• Trading or selling crypto for fiat currency
• Trading one crypto for another cryptocurrency
• Spending crypto to buy a good or serviceâ
That means that if youâre simply holding your cryptocurrency, you will not be able to deduct any losses. You will only be able to report your losses once a taxable event occurs.
You May Like: Where Do I Find Bitcoins
## While Currently There Are No Specific Guidance/specific Tax Provisions On Taxation Of Cryptos In The Income
cryptogainslossesReporting of cryptocurrency transactionscapital gainsTaxability under business income/capital gains
• Taxability as capital gains: If cryptos are held as investments, then it could be argued that the profit/loss on such sale needs to be reported as capital gains/loss. If the cryptos are held for more than 36 months, then the gain thereon could be classified as long-term capital gains and be subject to tax at 20%, plus applicable surcharge and cess. Else, they could be classified as short-term capital gains, subject to tax at the applicable personal taxation rates. For long-term capital gains, indexation benefit could be availed to increase the cost on account of inflation.
• Taxability as business income: If cryptos are held as stock-in-trade, then it could be taxed under the head business income. The income from such activity of trading could be taxed as business income. As mentioned above, for individuals having business income, the prescribed ITR Form, i.e., ITR-3 is to be used . Business income is taxed as per the prevailing slab rates , plus applicable surcharge and cess.
How to report in ITR-2/ITR-3Reporting of cryptocurrency holdings in ITRAdditional reporting requirement in ITRPenal consequences for not reporting cryptocurrencies in ITR
#### Read More News on
All you need to know about ITR filing for FY 2020-21.)
Latest news
Related news | 3,990 | 19,225 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-21 | longest | en | 0.944833 |
https://turtletoy.net/turtle/234952b55f | 1,726,501,933,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651697.45/warc/CC-MAIN-20240916144317-20240916174317-00592.warc.gz | 534,085,101 | 12,034 | ### Gradient path mutations 📿
In 8 directions! (or 4)
(...while maintaining cell size constraints)
```// Forked from "Bounded path mutation 📿" by Jurgen
// https://turtletoy.net/turtle/4e696524dd
// Forked from "Path mutation" by reinder
// Path mutation. Created by Reinder Nijhoff 2024 - @reindernijhoff
//
//
const evolution = 4; //min=4 max=8 step=4 (In 4 directions, In 8 directions)
const mutationRate = .195; // min=0.0, max=0.5, step=0.001
const mutationDefect = 57.3; // min=0, max=100, step=0.1
const mutationCountMin = 10; // min=1, max=100, step=1
const mutationCountMax = 30; // min=1, max=150, step=1
const grid = 13; // min=3, max=25, step=2
const pathInput = `M0,-37 C-10,-37 -26,-22 -30,-14 C-32,-10 -38,-3 -35,2 C-28,17 -7,39 13,29
C14,28 17,30 18,29 C59,9 37,-36 -3,-36`; // type=path, bbox=-40,-40,80,80 Click here to redraw the path
let seed = 1; // min=1, max=1000, step=1
let tokens = pathInput.match(/([0-9.-]+|[MLC])/g);
const populatedGrid = runUglyCodeToPopulateTheGrid();
function Translate(x,y) { return p => [p[0]+x, p[1]+y]; }
function Scale(s) { return p => [p[0]*s, p[1]*s]; }
function lerpTokens(a, b, p) {
return a.map((token, index) => {
if (isNumber(token)) {
return token * (1-p) + b[index] * p;
});
}
function mutation(tokens) {
if (isNumber(token)) {
return random() < mutationRate ? token - (random()-.5)*mutationDefect : token;
});
}
function walk(i) {
const y = i/grid|0, x = i%grid;
const path = Path(populatedGrid[x][y]);
const steps = path.length() | 0;
const turtle = new Tortoise(path.p(0));
turtle.addTransform(Scale( 132 / Math.max(...path.size()) / grid));
for (let i=0; i<steps; i++) {
turtle.goto(path.p( i/steps ));
}
return i < grid**2-1;
}
function isNumber(n) {
return !isNaN(parseFloat(n)) && isFinite(n);
}
function random() {
let r = 1103515245 * ((++seed >> 1) ^ seed);
r = 1103515245 * (r ^ (r>>3));
r = r ^ (r >> 16);
return r / 32768 % 1;
}
////////////////////////////////////////////////////////////////
// Modified path utility code. Created by Reinder Nijhoff 2023
// Parses a single SVG path (only M, C and L statements are
// supported). The p-method will return
// [...position, ...derivative] for a normalized point t.
//
//
// Modified by Jurgen Westerhof 2024, added bb() and size()
////////////////////////////////////////////////////////////////
function Path(tokens) {
class MoveTo {
constructor(p) { this.p0 = p; }
p(t, s) { return [...this.p0, 1, 0]; }
length() { return 0; }
}
class LineTo {
constructor(p0, p1) { this.p0 = p0, this.p1 = p1; }
p(t, s = 1) {
const nt = 1 - t, p0 = this.p0, p1 = this.p1;
return [
nt*p0[0] + t*p1[0],
nt*p0[1] + t*p1[1],
(p1[0] - p0[0]) * s,
(p1[1] - p0[1]) * s,
];
}
length() {
const p0 = this.p0, p1 = this.p1;
return Math.hypot(p0[0]-p1[0], p0[1]-p1[1]);
}
}
class BezierTo {
constructor(p0, c0, c1, p1) { this.p0 = p0, this.c0 = c0, this.c1 = c1, this.p1 = p1; }
p(t, s = 1) {
const nt = 1 - t, p0 = this.p0, c0 = this.c0, c1 = this.c1, p1 = this.p1;
return [
nt*nt*nt*p0[0] + 3*t*nt*nt*c0[0] + 3*t*t*nt*c1[0] + t*t*t*p1[0],
nt*nt*nt*p0[1] + 3*t*nt*nt*c0[1] + 3*t*t*nt*c1[1] + t*t*t*p1[1],
(3*nt*nt*(c0[0]-p0[0]) + 6*t*nt*(c1[0]-c0[0]) + 3*t*t*(p1[0]-c1[0])) * s,
(3*nt*nt*(c0[1]-p0[1]) + 6*t*nt*(c1[1]-c0[1]) + 3*t*t*(p1[1]-c1[1])) * s,
];
}
length() {
return this._length || (
this._length = Array.from({length:25}, (x, i) => this.p(i/25)).reduce(
(a,c,i,v) => i > 0 ? a + Math.hypot(c[0]-v[i-1][0], c[1]-v[i-1][1]) : a, 0));
}
}
class Path {
constructor(tokens) {
this.segments = [];
this.parsePath(tokens);
}
parsePath(t) {
for (let s, i=0; i<t.length;) {
switch (t[i++]) {
case 'M': this.add(new MoveTo(s=[t[i++],t[i++]]));
break;
case 'L': this.add(new LineTo(s, s=[t[i++],t[i++]]));
break;
case 'C': this.add(new BezierTo(s, [t[i++],t[i++]], [t[i++],t[i++]], s=[t[i++],t[i++]]));
break;
default: i++;
}
}
}
this.segments.push(segment);
this._length = 0;
this._bb = undefined;
this._size = undefined;
}
length() {
return this._length || (this._length = this.segments.reduce((a,c) => a + c.length(), 0));
}
bb(sampleRate = .01) {
if(this._bb === undefined) {
this._bb = Array.from({length: 1 / sampleRate + 1})
.map((v, i) => this.p(i * sampleRate))
.reduce((p, c) => [[Math.min(p[0][0], c[0]), Math.min(p[0][1], c[1])],[Math.max(p[1][0], c[0]), Math.max(p[1][1], c[1])]], [[Number.MAX_SAFE_INTEGER, Number.MAX_SAFE_INTEGER], [Number.MIN_SAFE_INTEGER, Number.MIN_SAFE_INTEGER]]);
}
return this._bb;
}
size(sampleRate = .01) {
if(this._size === undefined) {
this._size = [this.bb(sampleRate)].map(v => [v[1][0] - v[0][0], v[1][1] - v[0][1]]).pop();
}
return this._size;
}
p(t) {
t = Math.max(Math.min(t, 1), 0) * this.length();
for (let l=0, i=0, sl=0; i<this.segments.length; i++, l+=sl) {
sl = this.segments[i].length();
if (t > l && t <= l + sl) {
return this.segments[i].p((t-l)/sl, sl/this.length());
}
}
return this.segments[Math.min(1, this.segments.length-1)].p(0);
}
}
return new Path(tokens);
}
////////////////////////////////////////////////////////////////
// Tortoise utility code. Created by Reinder Nijhoff 2019
// https://turtletoy.net/turtle/102cbd7c4d
////////////////////////////////////////////////////////////////
function Tortoise(x, y) {
class Tortoise extends Turtle {
constructor(x, y) {
super(x, y);
this.ps = Array.isArray(x) ? [...x] : [x || 0, y || 0];
this.transforms = [];
}
this.transforms.push(t);
this.jump(this.ps);
return this;
}
applyTransforms(p) {
if (!this.transforms) return p;
let pt = [...p];
this.transforms.map(t => { pt = t(pt); });
return pt;
}
goto(x, y) {
const p = Array.isArray(x) ? [...x] : [x, y];
const pt = this.applyTransforms(p);
if (this.isdown() && (this.pt[0]-pt[0])**2 + (this.pt[1]-pt[1])**2 > 4) {
this.goto((this.ps[0]+p[0])/2, (this.ps[1]+p[1])/2);
this.goto(p);
} else {
super.goto(pt);
this.ps = p;
this.pt = pt;
}
}
position() { return this.ps; }
}
return new Tortoise(x,y);
}
// Way too ugly code stashed far away at the bottom so nobody will look at it
function runUglyCodeToPopulateTheGrid() {
const roundsMin = Math.min(mutationCountMin, mutationCountMax);
const roundsMax = Math.max(mutationCountMin, mutationCountMax);
const keyCells = [];
for(let i = 0; i < evolution; i++) {
let rounds = roundsMin + ((roundsMax - roundsMin) * random() | 0);
keyCells.push(Array.from({length: rounds}).reduce((p, c) => mutation(p), tokens));
}
if(evolution == 4) {
keyCells.splice(1, 0, lerpTokens(keyCells[0], keyCells[1], .5));
keyCells.splice(3, 0, lerpTokens(keyCells[0], keyCells[3], .5));
keyCells.splice(4, 0, lerpTokens(keyCells[2], keyCells[5], .5));
keyCells.splice(6, 0, lerpTokens(keyCells[5], keyCells[6], .5));
}
keyCells.splice(4, 0, tokens);
const populatedGrid = Array.from({length: grid}).map((v, c) => Array.from({length: grid}));
const beg = 0;
const mid = grid / 2 | 0;
const end = grid - 1;
[beg, mid, end].forEach(r => [beg, mid, end].forEach(c => populatedGrid[c][r] = keyCells.pop()));
for(let row = beg; row <= end; row += mid) {
for(let i = 0; i < 2; i++) {
for(let col = 1; col < mid; col++) {
populatedGrid[col + i * mid][row] = lerpTokens(
populatedGrid[i * mid][row],
populatedGrid[(i + 1) * mid][row],
col / mid
);
}
}
}
for(let col = beg; col <= end; col++) {
for(let i = 0; i < 2; i++) {
for(let row = 1; row < mid; row++) {
populatedGrid[col][row + i * mid] = lerpTokens(
populatedGrid[col][i * mid],
populatedGrid[col][(i + 1) * mid],
row / mid
);
}
}
}
return populatedGrid;
}``` | 2,647 | 7,438 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2024-38 | latest | en | 0.258089 |
http://www.dailytech.com/article.aspx?newsid=20247&commentid=637653&threshhold=1&red=2037 | 1,427,697,081,000,000,000 | text/html | crawl-data/CC-MAIN-2015-14/segments/1427131299114.73/warc/CC-MAIN-20150323172139-00066-ip-10-168-14-71.ec2.internal.warc.gz | 448,114,249 | 13,755 | Print 74 comment(s) - last by MCKENZIE1130.. on Nov 29 at 8:39 PM
Chevrolet Volt window sticker
93 mpg on battery power, 37 mpg on gasoline power
After a cluster bomb that was unleashed yesterday when the Nissan Leaf was rated for an EPA estimated 99 mpg -- even though it is a "battery only" vehicle -- General Motors is dropping a bunch of digits on us when it comes to the EPA rating for its Chevrolet Volt.
According to the window sticker that will be plastered on all new Volts sold in the U.S., the vehicle is rate at an equivalent of 93 mpg when running on electricity, and a more sedate 37 mpg when the gasoline engine kicks in after the battery is depleted. This two figures combined give the Volt a "composite" rating of 60 mpg.
And here are some more numbers -- the Volt will have an official "battery only" range of 35 miles, while the total driving range (taking into account the batteries and the gasoline tank) will be 379 miles.
When the Volt was first announced, GM said that the vehicle would have a 40-mile range when running on battery power. The company recently revised that figure to 25-50 miles.
The Volt will go on sale later this month with a price tag of \$41,000 before a \$7,500 federal tax credit.
Comments Threshold -1 0 1 2 3 4 5
By Solandri on 11/24/2010 7:26:56 PM , Rating: 4
Average car is driven 12000-15000 miles per year. Split the difference and call it 13,500 miles per year. Assuming it's a commuter car, it's driven 250 days a year, or 54 miles a day.
If you drive 54 miles in a day, 35 of which are on battery, 19 are on gasoline.
19 miles @ 37 mpg = (19 miles) / (37 miles/gallon) = 0.514 gallons
35 miles @ 93 mpg = (35 miles) / (93 miles/gallon) = 0.376 gallons
Total miles = 54. Total gallons = 0.890.
54 miles / 0.890 gallons = 60.7 mpg
"If you look at the last five years, if you look at what major innovations have occurred in computing technology, every single one of them came from AMD. Not a single innovation came from Intel." -- AMD CEO Hector Ruiz in 2007
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Copyright 2015 DailyTech LLC. - RSS Feed | Advertise | About Us | Ethics | FAQ | Terms, Conditions & Privacy Information | Kristopher Kubicki | 570 | 2,187 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2015-14 | longest | en | 0.921034 |
https://climateaudit.org/2015/07/03/ruling-out-high-deflation-scenarios/ | 1,685,419,729,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224645089.3/warc/CC-MAIN-20230530032334-20230530062334-00545.warc.gz | 214,255,316 | 114,748 | ## Ruling out high deflation scenarios
Further to my series of posts on Deflategate, reader chrimony observed that my statistical analysis had shown that it was possible that there had been no tampering, but had not excluded the possibility of tampering. This is a sensible observation, but raises the question of whether and how one could use the available statistical information to exclude tampering. This is analysis that ought to have been done in the Wells Report. I’ve done the analysis in this post and the results are sharper than I’d anticipated.
For Logo initialization, any manual deflation exceeding de minimis of say 0.1 psi can be excluded by observations. For Non-Logo initialization, statistical information rules out “high” deflation scenarios i.e. deflation by more than the inter-gauge bias of 0.38 psi plus uncertainty, including deflation levels of ~0.76 psi reported in Exponent’s deflation simulations. Remarkably, for Non-Logo initialization, the only manual deflation that is not precluded are amounts equal (within uncertainty) to the inter-gauge bias of ~0.38 psi. Precisely why Patriots would have deflated balls by an amount almost exactly equal to the bias between referee Anderson’s gauges is a bizarre coincidence, to say the least. I think that one can safely say that it is “more probable than not” that referee Anderson used the Logo gauge than that such an implausible coincidence.
As discussed in a previous post (see here), half-time ball pressures can be converted to ball temperatures using the Ideal Gas Law and knowledge and/or assumptions of pre-game initialization conditions.
The half-time ambient temperature was 48 deg F (black solid dot). The average Colt half-time pressures (using relatively unbiased Non-Logo gauge) convert to an average ball temperature of approximately 58.1 deg F (blue + sign). Based on the information that the referees only measured 4 Colt balls because they were “running out of time”, I’ve estimated the average Colt measurement time at 12.5 minutes, just before the end of half-time at 13.5 minutes. This yields the negative exponential transient as shown below. Because Patriots had substantially more ball possession, especially towards the end of the first half, their mix of balls would be wetter than the Colt mix and thus, if anything, below the Colt transient. Note that this transient is for a mix of wet and dry balls – not dry balls or wet balls.
Figure 1. Half-time ball temperatures. Dry transient is fitted negative exponential to Colt half-time average at estimated average measurement time of 8 minutes. Wet transient differential is based on information in Figure 27. Implied ball temperatures for average Patriot half-time pressure measurements is shown for three cases: Logo initialization and no deflation; Non-Logo initialization and no deflation; Non-Logo initialization and 0.72 psi deflation – matching average deflation in Exponent simulations of rapid deflation.
Exponent’s simulations of surreptitious deflation all yielded an average deflation of ~0.76 psi (with very little variability – see note in Appendix.) If Patriot balls had been deflated after measurement by the same amount as Exponent’s deflation simulation (0.76 psi) – a plausible comparison – then the implied ball temperature for Patriot half-time average pressure of 11.11 psi (Non-Logo gauge) is 58.4 deg F – higher than the corresponding temperature for Colt balls measured later in the half-time- and well above the transient at plausible average Patriot measurement times. The implied hiatus contradicts the possibility of surreptitious deflation in the amount of the Exponent simulations. The only deflation values (Non-Logo gauge initialization) that are consistent with the transient to observed Colt values are values in an interval centered (curiously) at ~0.38 psi – the precise value of the inter-gauge bias.
For comparison, I’ve also shown the corresponding ball temperature assuming Logo gauge initialization at 71 deg F (red + sign), almost exactly on the Colt temperature transient. If, after Logo gauge initialization, there had been manual deflation of 0.38 psi, the ball temperature corresponding to 11.11 psi at half-time would be almost exactly equal to the 0.76 psi deflation case for Non-Logo initialization – contradicted by the resulting hiatus.
Discussion
These results are considerably sharper than results in earlier discussion. For the Logo gauge initialization of Patriot balls, it is not just that observed values are consistent with Logo initialization, but any noticeable (in some sense) manual deflation would yield ball temperatures at 11.11 psi (Non-Logo) that were too high relative to the Colt measurements later in the half-time. Any manual deflation greater than ~0.1 psi or so would be inconsistent with observations. Exponent’s simulations did not show that such small deflation could be achieved, nor is there any sensible reason why anyone would bother trying to deflate footballs by ~0.1 psi.
For Non-Logo initialization, any manual deflation greater than ~0.38 psi plus uncertainty allowance, the observed half-time pressures of 11.11 psi (Non-Logo) yields ball temperatures that are too high in comparison with later Colt measurements and are precluded. Similarly, manual deflation less than ~0.38 psi minus uncertainty allowance yields ball temperatures that are too low in comparison with later Colt measurements.
The remarkable result is that only manual deflation (Non-Logo initialization) that is not precluded are amounts equal, within uncertainty, to the inter-gauge bias of ~0.38. It surely passes all understanding why the Patriots would set a deflation target that so exactly matched the inter-gauge bias of referee Anderson’s two gauges. And then executed a surreptitious deflation program exactly implementing this implausible objective. Or even why they would bother with ~0.38 psi deflation rather than more substantial deflation of 1-1.5 psi or more.
I think that one can safely say that it is “more probable than not” that referee Anderson used the Logo gauge than that such an implausible coincidence.
Appendix – Exponent’s Deflation Simulations
Exponent’s deflation simulations involved three different employees attempting to deflate 11 footballs in 1 minute 40 seconds. The results were very consistent: each employee deflated the balls by an average of 0.76 psi, with very narrow deflation ranges both intra-employee and intra-employee. The average deflation ranged from 0.75-0.79 psi and standard deviations of ~0.1 psi. Note that the Wells Report had arm-wavingly attributed Patriot pressure variability to variability in deflation, but their own deflation simulations did not yield anything other than negligible variability – an inconsistency not addressed in the report.
1. AnonyMoose
Posted Jul 3, 2015 at 2:56 PM | Permalink
I think that you’re talking about the NFL Deflategate, but you probably should explicitly define the context of each posting so we don’t misunderstand what you’re talking about.
Steve: done.
2. Posted Jul 3, 2015 at 5:21 PM | Permalink
Watergate, Climategate, Deflategate, it just gets better and better.
3. MikeN
Posted Jul 3, 2015 at 5:52 PM | Permalink
Exponent tried to deflate 12 footballs, I hope.
• MikeN
Posted Jul 4, 2015 at 4:00 AM | Permalink
Correction, there were 13 footballs in the Patriots bag.
• mpainter
Posted Jul 4, 2015 at 9:57 AM | Permalink
Why did Exponent run a deflation simulation with only eleven footballs ? This would change the feasibility conclusions, it seems.
• MikeN
Posted Jul 7, 2015 at 8:34 AM | Permalink
It’s a typo by Steve.
4. kim
Posted Jul 4, 2015 at 7:58 AM | Permalink
Round the Patriots’ Stars and Stripes,
Pop go the lies and tripe.
==========
• editstet
Posted Jul 4, 2015 at 2:43 PM | Permalink
Kim, So good to read from you. Life on the climate blogs is no good without you, or, at least, a lot more boring.
5. MikeN
Posted Jul 4, 2015 at 9:12 AM | Permalink
I wonder if this will be covered at the next Sloan Sports Conference.
6. Doug Reichlin
Posted Jul 4, 2015 at 9:51 AM | Permalink
So the Ref used the wrong gauge, and Brady takes the fall !?!?
While I want to believe you, it still doesn’t explain the “unusual behavior” of all involved (i.e. locking oneself and 13 balls in the bathroom!?!?).
Maybe the “balllboy’s” gauge was accurate (or inaccurate high!!), and they were just adjusting to “regulation minmimums” with an accurate/equally biased gauge, and offsetting the bias in the Ref’s incorrect gauge??
Unravel that one Sherlock! If anyone can, it’s Steve.
The real bottom line is that the “inflategate” did not affect the game’s outcome (whether real or not), and NFFL is scapegoating in response to (paying) public outrage at Brady.
• mpainter
Posted Jul 4, 2015 at 10:20 AM | Permalink
Concerning “locking himself” in the bathroom, I have not yet seen any real confirmation except “the video shows this”. Was there actually a camera in the restroom filming everyone as they took a whiz? Unbelievable!
But in fact, from one very real perspective, McNally had a very good and quite legitimate reason for locking himself in the bathroom with the balls, if he did so:
The game officials could not be trusted to not over inflate the balls, as in the Jets game where the Patriots’ balls were inflated to near 16 psi by the game officials. McNally would simply have gauged the balls to assure that they had not been over-pressurized by some negligent game official.
So there is more than one speculative interpretation of why McNally took the balls into the bathroom.
Of course, McNally may have only answered the call of nature.
Posted Jul 5, 2015 at 4:46 PM | Permalink
“Concerning “locking himself” in the bathroom, I have not yet seen any real confirmation”
It’s a matter or style and persuasion. McNally is the most likely–and only–source for the locked single toilet room door. But acceding such in the text would 1) show cooperation on McNally’s part and 2) detract from mystery-story angle of telling the tale of the tape. Further, acceding that McNally is the source of the bathroom lock information, and taking it to be true, undermines the report’s tacit suggestion we should treat as suspicious McNally’s mistaken recollection of a toilet to be “urinal.” (No joke!)
• Steve McIntyre
Posted Jul 5, 2015 at 8:05 PM | Permalink
I don’t understand how any conclusions can be drawn about locking the bathroom door. The Wells Report says that the bathroom in question had a single toilet:
The bathroom measures approximately 9 feet by 9 feet. It has a single toilet near the back right corner (if one is
standing in the doorway and facing into the bathroom), and a single sink directly across from the door.
I suspect that 99% of all Americans (and Canadians) would lock a single toilet bathroom when using it, just as they do at Starbucks. It amazes me that such drivel is considered relevant, while such a hash is made of the statistics.
• MikeN
Posted Jul 5, 2015 at 8:13 PM | Permalink
The Wells Report has the source to be the security video. I also don’t get why people are making a big deal out of it. I’m just curious how they knew it was locked from the video.
Posted Jul 5, 2015 at 9:39 PM | Permalink
Mike N.:
The Wells Report has the source to be the security video
If you look carefully, I don’t think that is exactly right. True, when discussing the bathroom, it is usually in proximity of discussion of a tape. But take a look at p. 4 para. 5:
Based on videotape evidence and witness interviews… … …McNally entered that bathroom with the game balls, locked the door…
Therefore, “interviews” were part of the evidence of the bathroom visit. Either that is McNally himself, or some guy who walked by and heard a lock click while looking at his wristwatch. It is a “big deal” because the crafting shows motive..of the Report’s wordsmiths.
Steve M.:
Page 20 has my fave: ..McNally’s disappearance into a locked bathroom.. He did not just enter a bathroom, he disappeared into it! I faintly hear Canadian Count Floyd (SCTV-MCHT) baying in him inimical way, “Ooooh, that’s scary!”
Steve: “Disappeared”?!? I wonder if they consulted the Wyndham’s Wizard (of the recent ads).
• Carrick
Posted Jul 6, 2015 at 11:12 AM | Permalink
Steve McIntyre:
I don’t understand how any conclusions can be drawn about locking the bathroom door. The Wells Report says that the bathroom in question had a single toilet:
You can conclude the balls, which were illegally in Deflator McNally’s possession, were now no longer visible.
Nothing suspicious there. No sir.
• Posted Jul 6, 2015 at 11:19 AM | Permalink
Carrick – “illegally”?
From a previous comment: “Also the Patriots ball attendant didn’t break any rules by taking the balls out of the waiting room. This was an unusual situation because the officials and teams were waiting for the overtime of the previous game to end. When the game ended one of the officials said “we’re back on”. That’s when the ball attendant took the balls out. Right in front of all the officials with no complaints. Then he stopped at the bathroom on the way to the field.”
• Carrick
Posted Jul 6, 2015 at 11:45 AM | Permalink
HaroldW, that’s pretty strange logic. It’s now legal because a prior infraction wasn’t penalized???
• Carrick
Posted Jul 6, 2015 at 12:01 PM | Permalink
To make sure we’re on the same page here:
The rule in question is Playing Rule 2, Section 1, which from Well’s report states:
Referee shall be the sole judge as to whether all balls offered for play comply with these specifications . . . and the balls shall remain under the supervision of the Referee until they are delivered to the ball attendant just prior to the start of the game
Possibly it is not an offense to carry the balls for the referees on McNally’s part (if so, it does not seem to be an enforced rule), but it is clearly a violation of Rule 2 to knowingly remove the balls from the supervisory controls of the officials.
Whether the bathroom is a single stall or otherwise, McNally taking them into a locked bathroom was a clear violation of Rule 2.
• Posted Jul 6, 2015 at 2:32 PM | Permalink
Carrick –
My understanding is that McNally was the ball attendant. So I don’t see a violation of “the balls shall remain under the supervision of the Referee until they are delivered to the ball attendant” here.
• Carrick
Posted Jul 6, 2015 at 9:02 PM | Permalink
HaroldW:
My understanding is that McNally was the ball attendant. So I don’t see a violation of “the balls shall remain under the supervision of the Referee until they are delivered to the ball attendant” here.
Huh?
McNally is an employee of the Patriots and obviously not an NFL official. How is McNally disappearing (heh) into a restroom consistent at all with the balls remaining in the supervision of the Referee?
Steve: Carrick, I think that the reader’s point was that McNally was the “ball attendant” for the Patriots within the meaning of the rule and it was just prior to the start of the game. Obviously the system was poorly constructed and most of us can agree on that. But, without the physical evidence, I’m not convinced that the evidence shows that it’s more probable than not that McNally tampered with the balls.
.
• chuckrr
Posted Jul 6, 2015 at 9:31 PM | Permalink
Carrick
“and the balls shall remain under the supervision of the Referee until they are delivered to the ball attendant just prior to the start of the game”
delivered to the attendant (The ref said “were back on”)….check
just prior to the game (15 to 20 minutes)…..check
What part of the rule was broken?
• Carrick
Posted Jul 6, 2015 at 10:40 PM | Permalink
chuckrr:
delivered to the attendant (The ref said “were back on”)….check
just prior to the game (15 to 20 minutes)…..check
What part of the rule was broken?
It needed to have remained in the control of the Referee until you reach the game field. Otherwise, as I’ve pointed out above, why bother letting the Referee even check at the footballs?
The point of the supervisory control is to prevent tampering. Allowing them to go out of sight and hence control of the Referee clearly is a violation of that principle.
• Posted Jul 6, 2015 at 11:38 PM | Permalink
Carrick –
Sorry for not being more clear. Steve M and chuckrr correctly explained my point of view. From your earlier comment, the rule reads, “the balls shall remain under the supervision of the Referee until they are delivered to the ball attendant just prior to the start of the game.” As chuckrr wrote, they were under the ref’s supervision, and then given to the ball attendant (viz., McNally) just prior to the start of the game. In accordance with the rule which you cited.
Your last comment adds something to that rule: “It [footballs] needed to have remained in the control of the Referee until you reach the game field.” But that isn’t the rule! It hardly seems fair to attach your interpretation of what the rule *should* say, and then blame the Patriots for not following it.
• Posted Jul 6, 2015 at 11:46 PM | Permalink
Carrick –
I see you’ve answered my questions in other comments, so no need to respond in this particular thread.
• chuckrr
Posted Jul 7, 2015 at 12:04 AM | Permalink
Harold’s point also enhances the absurdity of the whole bathroom deflation theory. Let’s say Carrick is right and the attendant was not supposed to take possession of the balls until he reached the field. We are required to believe that the attendant on his own or through orders by the Patriots took the incredible risk of purposely taking the balls out of the officials supervision….illegally. Wouldn’t it seem likely that if they did take that chance that they would have to have some foreknowledge of where the officials would be when he passed the bathroom. So if they knew where the officials would be then that would have to be the standard practice would it not? Or did the attendant just say to himself serendipitously…”here’s my chance…go for it” I’m really trying to come up with a realistic scenario…somebody help me out.
• Carrick
Posted Jul 7, 2015 at 12:17 AM | Permalink
chuckrr:
We are required to believe that the attendant on his own or through orders by the Patriots took the incredible risk of purposely taking the balls out of the officials supervision….illegally. Wouldn’t it seem likely that if they did take that chance that they would have to have some foreknowledge of where the officials would be when he passed the bathroom.
This is addressed in the Well’s report too. It appears that on normal game days, there would have been space in the locker room (which is big), for somebody to have tampered with the footballs without being noticed. So what McNally did during that playoff game was unusual (removing the footballs without permission and/or supervision), and he possibly felt forced into taking a bigger risk that he normally might have.
Anyway, based on your logic, crime would be virtually non-existent in our world. Unfortunately, most people who engage in illegal behavior also take foolish chances so your logic doesn’t follow.
• chuckrr
Posted Jul 7, 2015 at 1:09 AM | Permalink
As with everything else in the Wells report it proves nothing. ….. We have no evidence past balls were tampered with but we do know he had an opportunity to tamper with the balls in past games that we have no evidence he tampered with….. My point was not that criminals don’t take chances and people don’t do stupid things.My point is simply what makes the most sense. To you and the Wells proprietorial team obviously it makes sense. I think we’ve seen this kind of proprietorial assumption of guilt many times. When your convinced of guilt everything seems to prove your point.Is it possible…sure. Don’t take this wrong I know your analytical abilities far exceed mine. And your unbiased as well. But we are all human
• chuckrr
Posted Jul 7, 2015 at 1:13 AM | Permalink
opps…proprietorial is supposed to be prosecutorial …ruined a brilliant comment
• MikeN
Posted Jul 7, 2015 at 8:36 AM | Permalink
Chuck, if the whole thing can be done in under 90 sec as Exponent suggests, then McNally would have been better off doing so in the locker room, despite how crowded it is.
• Carrick
Posted Jul 7, 2015 at 9:03 AM | Permalink
checker:
We have no evidence past balls were tampered with but we do know he had an opportunity to tamper with the balls in past games that we have no evidence he tampered with
Um no. Page 44 of the Wells report.
Game 11 with Indy. Two interceptions by Mike Adams. The Colts noticed that both footballs seemed under inflated. The Indy GM sent an email to the league prior to the game warning about the Patriots possibly tampering with the footballs.
MikeN:
Chuck, if the whole thing can be done in under 90 sec as Exponent suggests, then McNally would have been better off doing so in the locker room, despite how crowded it is.
I think the theory is McNally normally deflated the balls in the locker room, but it was packed that day: The Wells report suggests it would have been impossible to deflate the balls without being seen, so what you are suggesting wouldn’t work for that playoff game but in general would have been true.
• MikeN
Posted Jul 7, 2015 at 11:44 AM | Permalink
The Colts game with the two interceptions was at Indianapolis, so the deflator would have had no chance to use his needle. However the temperature was 28-35 degrees.
• chuckrr
Posted Jul 7, 2015 at 7:57 PM | Permalink
Carrick
An accusation (the other colt game) is not evidence. It was cold and the Colts are obviously looking for excuses
Posted Jul 5, 2015 at 9:07 PM | Permalink
I’ll relate one more “story-telling” effect of mysterious goings-on deep inside stadiums: the omission of the halftime’ refs statements regarding the logo/non-logo switcheroo. In short, at halftime refs Blakeman and Prioleau, using two different gauges (“non-logo” and “logo”), tested 11 Patriots’ balls and 4 Colts balls. There were four other witnesses. The Patriots’ balls were tested first. Alarmingly, the two gauges showed significantly different readings. In a normal world this would put at question both gauges validity for any analytical use by the NFL at all, let alone a basis to charge high fines and cause job loss. But the report found the readings to be “consistent.” At page 67 the report says Blakeman and Prioleau, “reported no difficulties in using the gauges.”
At page 68 lies a table for the 11 Patriots’ ball tested. Each of Prioleau’s readings is higher than Blakeman’s. The readings range from .3 to .45 psi in difference.
At page 69 lies the table for the 4 Colts’ balls tested. Again each of the readings of Prioleau and Blakeman differ. These readings range from .35 to .45 in difference. However, this time 3 of Prioleau’s 4 readings are lower than Blakeman’s. The other reading of Prioleau is higher, like his Patriot balls’ measurements.
This evident problem is admitted at footnote 41, “it appears most likely that the two officials switched gauges in between measuring each teams footballs.” Why would Prioleau and Blakeman do that? What reasons did they give? Any? Wait…the basis for the ascertainment of the switching is the report’s opinion that one ref’s readings are consistently different than the other’s. As for the fourth non-conforming reading of a Colt’s football psi, that is attributed to a speculated clerical error. OK, maybe this all happened. What did Prioleau and Blakeman say about it? Did they admit the switch? Nothing in the report. Did they deny the switch? Nothing in the report. What about the witnesses, did they see the switch? Not in the report. Did they say that such would have been impossible? Not a thing in the report.
Now it would strain credulity that the six men in the room were not asked about the potential logo/non-logo switcheroo. That such critical information is absent, not even buried in a footnote, tells me that likely the officials’ accounts declined the possibility of a switch. Now, the switch may have happened anyway, somehow, but if this were an impartial report such information would be declared. But this report is prosecutorial “story-telling,” with the probable omission of on-the-scene, eyewitness evidence that does not help the conclusion the report seeks. Like the probability of McNally himself being the source of the locked door data, evidence that does not help the prosecution’s story-line of casting suspicions is omitted. That’s fine in a courtroom, but here this report purports to be an independent investigation. I think the ref and eyewitness statements on these issues must be revealed or this report be tanked as crud from these described omissions alone.
• mpainter
Posted Jul 5, 2015 at 9:51 PM | Permalink
Some good points are made by followthemoney.
Concerning the variance in the gauge readings between the the two refs Blakeman and Proileau compounded by inconsistencies (switched gauges?; clerical error?). Adding to the mess the lack of ball temperature data, then the _only_ conclusion that can reasonably be drawn from the ball pressure data is that _no_ conclusion can be drawn.
Steve: the gauge inconsistency was discussed at length in my original writeup. I do not agree that this makes everything unusable since one of each pair of measurements was about 0.38 psi higher than the other and it is known that the Logo bias was about 0.38 psi. Thus, it is, in my opinion, reasonable to attribute the inconsistencies to an inadvertent exchange of gauges and a transcription error for who measured the third Colt ball – as done in the Wells Report. On the other hand, this inattentiveness supports the idea that referee Anderson had inattentively made the same exchange in his pregame measurements, using the Logo gauge for Patriot balls and Non-Logo gauge for Colt balls. Please see earlier discussion and my original writeup for details.
.
• MikeN
Posted Jul 6, 2015 at 8:58 AM | Permalink
This is more evidence that the Patriots balls were inflated before the Colts balls were measured.
• mpainter
Posted Jul 6, 2015 at 12:42 PM | Permalink
Steve,
Having re-read your analysis, I agree that reasonable assumptions can be made regarding the use of the logo and non logo gauges and that reasonable inferences might follow.
• chuckrr
Posted Jul 4, 2015 at 10:31 AM | Permalink
“unusual behavior” What if the guy had to relieve himself as he was taking the balls to the field. I know that is very unusual but it is possible..right? Would he leave the balls outside the bathroom unattended? That would be less unusual behavior? And if in the incredibly unlikely scenario that he was using the bathroom to relieve himself would it be that unusual that he would lock the door? i guess so. What this incident really shows me is how the human mind can take any set of facts and interpret them to fit a narrative.
• Tom O
Posted Jul 6, 2015 at 1:02 PM | Permalink
A question about the “locking himself in the restroom.” We ARE talking about someone that has worked at the facility for years, are we not, and therefore is fully aware of all the security cameras? Seriously? You’re going to drag a bag full of footballs into a restroom that you KNOW is watched by a security camera, thus you are leaving evidence of what you doing in spite of the fact you are doing something wrong? I’ll bet you believe the Mann hockey stick, too.
• Carrick
Posted Jul 6, 2015 at 1:19 PM | Permalink
At this point, there’s no way to establish whether McNally knew about the video camera or not.
As to point of fact, McNally did enter the restroom, he did lock the bathroom, he did violate Rule 2 by removing the footballs from the supervision of the Referee, his actions were caught on camera and he did remain in the bathroom long enough to deflate all 13 footballs.
These are facts, what you are engaged in is just pointless speculation.
• chuckrr
Posted Jul 6, 2015 at 3:00 PM | Permalink
Carrick,
“Under the supervision of” is a pretty vague and general term don’t you think? Does that mean the ref is constantly watching the balls…uninterrupted. Is there an official phrase that the ref gives “officially” instructing the ball attendant to move the balls to the field? Does the ball attendant have to follow a predetermined official NFL route to the field? Are bathroom stops only allowed when permission is given by the official? I suspect the last thing the officials are thinking about after they check the their pressure is the footballs. And I think when the ref says “lets go” the guys carrying the balls take them out to the field…and even stop at the bathroom once in awhile.
• mpainter
Posted Jul 6, 2015 at 4:08 PM | Permalink
Does the Wells Report say that McNally violated protocol in this respect?
If it does not, that should settle the issue.
• MikeN
Posted Jul 6, 2015 at 4:36 PM | Permalink
“until they are delivered to the ball attendant.” You quote the rule and then ignore it. Indeed, the referee told him to bring the balls to the other room and then game start was postponed. So the referee had also delivered the balls prior to the start of the game, as per the rule.
Later in the rules it states the balls are not to be tampered with. That is the focus.
• Carrick
Posted Jul 6, 2015 at 10:08 PM | Permalink
chuckr, I don’t think there’s anything particularly vague about this. McNally taking them into the restroom with him certainly violates any plausible requirement for supervision. The purpose of the supervision is to prevent the balls from being tampered with until they reach the field of play (at that point presumably there are enough eyes and cameras to preclude further attempts at tampering).
mpainter:
Does the Wells Report say that McNally violated protocol in this respect?
Yes, they do say McNally violated protocol on this. They come pretty close to saying he lied about doing so too. That was one of the reasons they wanted a follow up interview with him (and probably why the Patriots refused to allow it).
MikeN:
“until they are delivered to the ball attendant.” You quote the rule and then ignore it. Indeed, the referee told him to bring the balls to the other room and then game start was postponed. So the referee had also delivered the balls prior to the start of the game, as per the rule.
If you read the text of the Well’s report, it’s pretty clear the delivery to the ball attendant occurs on the field of play, and not in the locker room.
What’s the point of having the Referee confirm that the footballs are within their permissible range, if you then allow the teams to handle them for an indeterminate amount of time with no supervision?
• chuckrr
Posted Jul 6, 2015 at 11:09 PM | Permalink
Carrick,
This is something I’m probably never going to convince you of but I’m pretty sure that prior to this incident the control of the balls was pretty lax. And the refs didn’t carry the balls to the field. This game was a little unusual as it was contingent on the previous playoff game finishing before it could start. Normally they start at a specified time. Another thing…and this is just a personal observation. Take it for what it’s worth. I’ve been an athlete my entire life. Played several sports including in college. Now I haven’t played professionally but I know and have been around NBA players and have known refs at that level and see how they think. My observation is that they don’t care about the balls because up until this incident there have been almost zero issues with the balls. They are not going to worry about a rule that nobody ever questions and few people consider important. They get judged on how they make an interference call and are they consistent. I would bet that no ref has ever been reprimanded for the way they transferred possession of the balls to the ball attendant.
• Carrick
Posted Jul 7, 2015 at 12:08 AM | Permalink
chuckrr, if you read the Wells report, you’ll see that the response of the Referee was anything but relaxed when the footballs came up missing. Nor was it standard procedure to allow the ball attendant to remove the footballs from the locker room and bring them to the game area without supervision:
According to Anderson and other members of the officiating crew for the AFC Championship Game, the removal of the game balls from the Officials Locker Room by McNally without the permission of the referee or another game official was a breach of standard operating pre-game procedure. According to Anderson, other members of the officiating crew for the AFC Championship Game and other game officials with recent experience at Gillette Stadium, McNally had not previously removed game balls from the Officials Locker Room and taken them to the field without either receiving permission from the game officials or being accompanied by one or more officials.
• chuckrr
Posted Jul 7, 2015 at 12:34 AM | Permalink
Carrick, An as I said this was an unusual game in that they were waiting for the overtime of the previous game to finish. And when the ref said “were back on” McNally thought he had permission to go. I suspect that the refs didn’t think twice about where McNally was until later when the investigators pointed out what happened. Then they said..”Yeah that was unusual”. You have to remember that this report was put together as a prosecutor would present his case. We see it over and over in the report…as other commenters have pointed out. Regardless Did the Patriots tell McNally to break out of there and get to that bathroom even at the risk of some official saying…where is he going? And how were they so sure the official wouldn’t follow him? Did the have spies with disposable phones? I’m sorry it seem so unlikely to me..not impossible but very unlikely
• Carrick
Posted Jul 7, 2015 at 12:52 AM | Permalink
chuckrr, based on the testimony there was definitely concern, possibly to the point of panic when the game balls could not be located:
When the remaining officials walked into the sitting room area on their way to the field, all four were surprised to find that the ball bags were not there. Both Anderson and Veteri immediately asked Farley where the footballs were. Farley checked for the ball bags in the back part of the locker room (where he saw the bags of back-up balls) and in the adjacent Chain Gang Locker Room, but could not find them. When it was suggested that McNally had or may have taken them to the field, Anderson responded that “he‟s not supposed to do that.” Anderson also stated that “we have to find the footballs.” Blakeman recalls that although Anderson is usually calm and composed leading up to a game, Anderson was visibly concerned and uncharacteristically used an expletive when the game balls could not be located. The other officials were similarly surprised and concerned. None of the officials in the locker room at the time realized that the game balls had been removed from the locker room until they were ready to go to the field for the start of the game, and all expected that the balls would not leave the locker room until it was time for them to take the field.
This concern appears documented by on-field video:
Although the officials were concerned about the situation, with kickoff approaching, they decided to take the field. Farley and the officials left the Officials Locker Room and walked to the field at approximately 6:36 p.m. As seen on the security footage, Farley walked approximately 10 seconds ahead of the officials because, as he explained, he was in a hurry to reach the field to look for the footballs. As soon as he reached the field, Farley looked for McNally by the instant replay booth, where McNally regularly arrives with the game balls, but did not see him. He did, however, see John Raucci, Director of Investigative Services at the NFL, shortly after stepping onto the field and asked if Raucci had seen either McNally or the game balls. Raucci said that he had seen neither. In an effort to ensure that the teams had footballs on the field for the start of the game, Farley headed back toward the Officials Locker Room to get the back-up balls. He is seen on the security footage at approximately 6:42 p.m. walking back down the tunnel leading to the field with the bags of back-up balls. Farley reported that prior to the AFC Championship Game, he has never been in a situation where the game balls could not be located or where he had to retrieve the back-up balls from the Officials Locker Room prior to kickoff.
This doesn’t seem consistent with a concern manufactured after the fact. To summarize:
Only based on the craziest possible reading of the rules can you come to the conclusion it would be in any manner appropriate for McNally to take the footballs. (This entirely negates the whole point of verifying the footballs initial inflation, since McNally could then tamper with BOTH teams footballs.)
It does not seem that McNally taking the footballs was at all standard practice. Based on the statement of the officials, this is the first time he had ever done this.
It does seem that the officials were agitated to the point that this could be seen by their on-field behavior.
Nobody has claimed that the Patriots ordered McNally to take the balls from the locker room (as this is the first time this has happened, it is not even consistent with his standard MO for deflating footballs, presuming he did so).
Nor am I claiming that the Patriots leadership were initially aware of any supposed impropriety (I think they were unaware). I don’t think the Wells report claims that either.
The biggest slam against the Patriots leadership was a failure to fully cooperate. And that’s certainly the case, regardless of what motivation you want to paint on why they didn’t fully cooperate.
• MikeN
Posted Jul 7, 2015 at 8:59 AM | Permalink
>McNally taking them into the restroom with him certainly violates any plausible requirement for supervision.
Until delivered to the ball attendant. There is no requirement that they continue to supervise.
>presumably there are enough eyes and cameras to preclude further attempts at tampering).
The Patriots rebuttal mentions that Colts were reported by the Jaguars as having needles under their sleeves. If each one can be done in close to saying he lied about doing so too.
Not close, of course they are calling him a liar.
>delivery to the ball attendant occurs on the field of play, and not in
The referees don’t carry the balls out. After the fact they claimed they always walk with him down to the field.
>What’s the point of having the Referee confirm that the footballs are within their permissible range, if you then allow the teams to handle them for an indeterminate amount of time with no supervision?
The rules say you are not to tamper with them afterwards. They do not say that the referee should supervise them all the way to the field. They might change this with a notice, but they can’t police everything.
• MikeN
Posted Jul 7, 2015 at 9:13 AM | Permalink
There is a contradiction between the Wells Report and the Pats rebuttal which claims that the video shows McNally coming out of the tunnel and going to the area next to the replay booth, which is where Mr Anderson found him.
• chuckrr
Posted Jul 7, 2015 at 8:34 PM | Permalink
Carrick,
“based on the testimony there was definitely concern, possibly to the point of panic when the game balls could not be located:”
Panic…really…. with that kind of hyperbole I’m starting to think your one of the authors of the Wells report
Anderson admitted that sometimes the he did not maintain supervision of the balls as he would have some other task to do. So the notion that the balls are always under the officials supervision is provably false from…the Wells Report. McNally with the backing of witnesses has said that about 50% of the time he took the balls out to the field unattended
Also from the Wells report…
“McNally had not previously removed game balls from the Officials Locker Room and taken them to the field without either receiving permission from the game officials or being accompanied by one or more officials.”
Take notice the part about giving permission or being accompanied. It doesn’t say always accompanied. McNally thought he had permission. Anderson didn’t think he had given permission. In a normal game Anderson probably wouldn’t have given that a second thought. But guess what, Andersons view of normal occurrences had been altered by the fact that he had been warned to look out for something. Things he probably would never notice suddenly become unusual. I’ve witnessed this phenomenon countless times with refs and everyday life. Why do you think coaches and players work refs. You can alter the way the ref sees the game.
And despite being warned the refs were still incredibly lackadaisical in their ball supervision, What does that say about their normal supervision? And what does that say about their claim that this situation was unusual? Is it possible that they were covering their own posteriors?
• MikeN
Posted Jul 7, 2015 at 9:07 AM | Permalink
Carrick, the security video shows McNally walking with the footballs unaccompanied, right by one of the officials who was notified before the game of the Colts’ concern with deflation. He made no objection to this guy taking the footballs by himself.
Even after they reinflated the Patriots balls at halftime, the league officials still had McNally take the balls to the field unaccompanied by anyone. The ‘he’s not supposed to do that’ was perhaps because they hadn’t specifically told him yet, but the idea that he is always accompanied is either bad memory or just after the fact argument.
Sure they couldn’t find the footballs so they went looking. Once they found them, no one criticized McNally for taking them by himself.
This was in a game where beforehand the referees were on notice about the footballs. If it was so unusual, perhaps he would have ordered the backup balls to be used, and for the originals to be tested.
• Carrick
Posted Jul 7, 2015 at 9:48 AM | Permalink
MikeN:
Carrick, the security video shows McNally walking with the footballs unaccompanied, right by one of the officials who was notified before the game of the Colts’ concern with deflation. He made no objection to this guy taking the footballs by himself.
On the field, initially nobody knew where McNally was. But it was Farley, not Anderson, who would couldn’t locate McNally and re-entered the locker room looking for the back up balls. And the issue appears to be whether the balls remain under supervision of the Referee not whether McNally was carrying them.
I think Anderson’s concern, as reported by Wells, is that McNally left the locker room without an escort. I don’t see any dispute that McNally took the footballs without being asked either, or that this lead to documented confusion on the field of play.
• MikeN
Posted Jul 7, 2015 at 10:06 AM | Permalink
Yes but if McNally went right there according to the Patriots, then why didn’t Farley see him?
• Carrick
Posted Jul 8, 2015 at 2:32 AM | Permalink
I don’t think it’s really in dispute that Farley did not see McNally. I don’t think the video evidence shows Farley walking right by McNally either. Am I wrong about that?
The Patriots did not claim Farley walked right by McNally either.
So it doesn’t seem to be a fair characterization to say “McNally was right there” at the point that Farley reentered the stadium tunnel.
• MikeN
Posted Jul 8, 2015 at 1:47 PM | Permalink
The Patriots claim that McNally, according to the video, left the bathroom and went straight to the area where everyone expected him to go. The Wells Report says Farley looked for McNally there, didn’t see him there, and went to get the alternate footballs. Those strike me as contradictory statements.
• Kevin
Posted Jul 11, 2015 at 2:43 PM | Permalink
Actually 12 balls were used in the first half.
During a bad weather game each team submits 24 balls, 12 are brought to the field 12 are left in the ref’s locker room in case they are needed.
Ironically the NFL destroyed the evidence. The could have impounded the ‘questionable’ balls at halftime and substituted the backup balls Anderson had approved that no one else had possession of.
This would have allowed the balls to be measured after the balls had dried and returned to room temperature.
Based on comments Steve Kensil, NFL executive he had no idea that there was a Ideal Gas Law, he told a Patriots employee they were in ‘big explective trouble’ because the balls underweight. 😀
• MikeN
Posted Jul 11, 2015 at 5:46 PM | Permalink
No it was 13. 12 is the usual number.
7. Shawn Marshall
Posted Jul 5, 2015 at 8:42 AM | Permalink
you got a lot of balls Steve.
8. chrimony
Posted Jul 5, 2015 at 10:07 AM | Permalink
I should weigh in here, since I was mentioned by name in the post. My previous position was: “In the face of inconclusive statistical analysis versus damning (in my opinion) evidence of tampering via the texts, video, and eyewitness accounts, I go with the latter.”
Now that Steve has done an analysis that shows it implausible that the Patriots intentionally deflated footballs, and that environmental factors combined with use of a particular gauge is the plausible reason for the measurements, I have two contradictory arms of evidence. How to reconcile them?
If I were the commissioner and had to decide guilt, without a convincing response to Steve’s analysis I would have to lean towards the presumption of innocence and not rule against them. Note that I still think something fishy is going on, and I find the explanations for the texts, video, and eyewitness accounts implausible. However, there, is no clear, direct evidence that the Patriots tampered with the balls in the Colts game or any other game for that matter, even though the texts strongly suggest it, there was motivation to do so, and distinct opportunities to do so.
• Joe
Posted Jul 5, 2015 at 2:50 PM | Permalink
Chrimony – mpainter makes reference to the only documentation of the balls being taken into a restroom was ” the video shows the ball boy going into the restroom with the balls”
Is there any actual documentation to this event? Or does it remain “someone said there is a video of him entering the bathroom?
Concerning “locking himself” in the bathroom, I have not yet seen any real confirmation except “the video shows this”. Was there actually a camera in the restroom filming everyone as they took a whiz?
• joe
Posted Jul 6, 2015 at 3:57 PM | Permalink
I stand corrected on the evidence of Mcnally entering the bathroom with the game balls. Security camera captured the event. Page 57/58 of the wells report.
The wells report makes reference to one minute & 40 secs (100 seconds) that mcnally spent in the bathroom. Lets assume for arguments sake that McNally did deflate the balls. At half time, all the balls were deflated by approx 1.0-1.5psi (taking into account the lower temps and therefore lower psi at halftime). There was one ball at 10.9 psi. (ie 1.6psi under inflated.)
To remove the 12 balls (11 balls?) from the ball bag, deflate each ball by approx .5-.6 psi, ensure that no balls were excessively deflated below that .5-.6 threshold, put all the balls back into the ball bag in those 100 seconds (approximately 8-9 secs per ball) requires a lot of skill and coordnation. I would suspect that it could be done in 80-90 secs by someone experienced. Whether McNally had the necessary skill was not addressed in the Wells Report.
It should be noted that McNally likely did not have a pump with him to reinflate the ball if he had underinflated any of the balls the “preferred threshold”.
• MikeN
Posted Jul 6, 2015 at 4:38 PM | Permalink
The bag actually had 13 balls. The Patriots’ rebuttal points out that the evidence that the deflation could be done was dated the same day as the release. It looks like they were not testing to see if it could be done, but testing to say it could be done.
• Steve McIntyre
Posted Jul 6, 2015 at 9:32 PM | Permalink
One of the 13 balls was the intercepted ball which the NFL kept in its possession. The NFL should have measured the pressure in this ball once it returned to equilibrium – even the following day. Like Jastremski did with the balls in the Jets game. My guess is that, at equilibrium, it would have measured 12.5 psi on the Logo gauge and 12.12 on the Non-Logo gauge.
• Joe
Posted Jul 6, 2015 at 8:36 PM | Permalink
MikeN – with 13 balls, that is approx 7.5 secs per ball, take the balls out of the bag, to deflate each one by .5psi-.6psi, put all the balls back in the bag, all the while making sure you dont excessively deflate any of the balls (since he likely did not have pump to correct any of over-deflation), then flush the toilet in order to create the alibi. A skilled person working efficiently could probably do it in 85-90 secs, leaving very little time for error. However, that would be a skilled working efficiently with no hiccups in the process. Amazingly, each ball was only deflated by .5-.6 psi – that takes a lot of talent.
The second point which others have alluded to is that are several security camera’s. The bathroom was apparently in a main hallway leading to the field ( and reasonably close to the field entrance between locker rooms and the field entrance). If that is the correct location, there would be a lot of traffic. McNally would have to be a DA (DA is not district attorney) to use that location for the deflation – assuming he did deflate the balls.
• chuckrr
Posted Jul 6, 2015 at 9:52 PM | Permalink
Joe,
From the beginning this bathroom thing seemed preposterous. And the more I learn the more unbelievable it seems. But others…probably much smarter than me… see it exactly the opposite. I guess it says a lot about confirmation bias. MikeN’s link somewhere here is an excellent explanation of this.
• Carrick
Posted Jul 6, 2015 at 10:21 PM | Permalink
Joe–it has been tested this, and shown it’s possible to have deflated 13 balls in approximately that amount of time. Since the amount of deflation is directly proportional to the amount of time its being deflated, that is the simplest part to explain.
Certainly he was aware of the consequences for him and his team if he were caught, so presumably if McNally were involved in illegal tampering, he would have been well practiced ahead of time.
• Steve McIntyre
Posted Jul 6, 2015 at 10:56 PM | Permalink
shown it’s possible to have deflated 13 balls in approximately that amount of time
I don’t think that the deflation experiments have had enough attention paid to them. With minimal practice, three different Exponent employees each deflated the footballs by approximately 0.76 psi with very small standard deviations between the three employees and between balls. Elsewhere the Wells Report speculated that the high variation in observed pressure of Patriot balls could only result from manual tampering, but their own simulations showed that even novice deflators could achieve consistent results – in other words, some other explanation is required for the high inter-ball pressure variability of PAtriot balls. Differing wetness seems the most plausible explanation to me, but this in turn requires analysis of evaporative cooling – a topic ignored by Wells and Exponent, even though they empirically observed lower pressures in wet balls without change in volume.
In addition, the experiments all resulted in ~0.76 psi deflation, an amount of deflation that (notwithstanding Carrick’s remarks) seems as inconsistent with observations as anything that the Wells Report objected to.
• Carrick
Posted Jul 7, 2015 at 12:25 AM | Permalink
Steve McIntyre:
I don’t think that the deflation experiments have had enough attention paid to them.
I don’t know why there needs to be a huge amount of attention here.
With a bit of practice and a choice of needle gauges, just about anybody could achieve pretty decent repeatability.
Elsewhere the Wells Report speculated that the high variation in observed pressure of Patriot balls could only result from manual tampering, but their own simulations showed that even novice deflators could achieve consistent results – in other words, some other explanation is required for the high inter-ball pressure variability of PAtriot balls.
I don’t buy Well’s argument here, but this is not particularly inconsistent—Assuming they were tampered with, you don’t know the order that the footballs were tampered with. If McNally was progressively letting less air out of the ball (suppose he felt progressively more rushed) and you sampled the balls later in random order, what was originally a smooth trend would look like a large amount of scatter.
• Steve McIntyre
Posted Jul 5, 2015 at 3:16 PM | Permalink
I should weigh in here, since I was mentioned by name in the post.
• Steve McIntyre
Posted Jul 5, 2015 at 3:52 PM | Permalink
I find the explanations for the texts, video, and eyewitness accounts implausible.
I’ve thought about doing a detailed exegesis of the texts, but it takes time to write up in the thorough way that I do.
Breaking the texts into the following pieces: (1) the texts after the Jets’ game; (2) the texts in December 2014-Jan 2015; (3) the early 2014 texts and at the risk of presenting views without fully backing up the interpretation.
McNally’s other texts in group (1) remind me of an overlooked junior employee complaints about the company golden boy. Also given the context that Brady was angry about referee over-inflation in violation of the rules, I don’t see anything in these other texts to suggest anything other than Brady insisting on his right to have 12.5 psi without interference by the referees, with McNally making idle threats.
As to group (2), nor do I find it odd or troubling that Brady would spend time talking to Jastremski after the controversy blew up. I don’t think that you can read anything into that one way or the other.
The “Deflator” text is troubling. McNally’s statement that it referred to weight loss seems Nick Stokesian, as (I think) you pointed out. But even Stokes is not always wrong in such parsing. (The issue with Stokes is rather his unwillingness to concede an inch when he is wrong on his parsing or his parsing is even more implausible than the text.) While Stokesian, McNally’s explanation is not necessarily untrue either. It seems to me that it has more weight as supporting evidence to “scientific” evidence of Game Day inflation than as the centerpiece of the case.
• chuckrr
Posted Jul 5, 2015 at 9:29 PM | Permalink
It should be noted that the term deflator was used only once and that was on May 9th of the previous year during the off season.
• Carrick
Posted Jul 6, 2015 at 2:07 PM | Permalink
chrimony:
Now that Steve has done an analysis that shows it implausible that the Patriots intentionally deflated footballs, and that environmental factors combined with use of a particular gauge is the plausible reason for the measurements, I have two contradictory arms of evidence. How to reconcile them?
I don’t agree the conclusion that it is implausible that the Patriots balls were more deflated than the Colts balls.
IMO Steve’s shown that, with enough untestable ad hoc assumptions thrown into an already muddy soup, you can construct plausible alternative theories where the Patriots balls weren’t more deflated. Unfortunately, Steve has also engaged in “fantasy football physics” in arriving at his conclusions—Based on my research experience, the assumption of exponential warming is unlikely to be correct here, and so shouldn’t be used to disallow alternative hypotheses or to construct new hypotheses (not without empirical measurements that mimic the actual ball handling process). Nor can you assume that the two set of footballs will follow the same warming curve, exponential or otherwise. This is because of differences in handling: The main driver for warming is convective air exchange. When the balls are in the ball bags the warming curve is going to look very differently than when they are removed.
I would also say that the Well’s report used similar “fantasy football physics” to arrive at the conclusion that the evidence was consistent with the Patriots balls being more deflated. I don’t see this as problematic as long as the purpose of the physics analysis is merely to demonstrate that the objective measurements are consistent with the other, frankly more exculpatory evidence that chrimony mentions.
It is my personal speculate that the Patriots did not intend to under-inflate the balls, rather just to reduce them to the lower end of the permissible range. I’d further say that some ball handling by the team to optimize the football pressure for their team should have been and should be permitted, so I think that a rule change similar to the 2006 Brady-Manning rule change is in order.
Anyway, I do not think the physics data (including metadata such as how the balls were handled before being measured) are good enough to say “this hypothesis is plausible but that one is not”. Were we left up only to the physical evidence, certainly I do not think any disciplinary action by the league would have been appropriate. Due to handling errors, I also don’t think the physical evidence is never going to be good enough to rule out the alternative plausible hypotheses.
I still think it the most parsimonious explanation based on all of the evidence is that the Patriots balls were simply inflated to a lower initial pressure, and that this was done deliberately by the Patriots staff, more likely than not in collusion with Tom Brady.
But as we all know, but some may be less willing to acknowledge, the amount of cooperation between the New England Patriots and the league was very limited (only a single interview with McNally was permitted, and no followup interviews were allowed) and the contractually-obligated cooperation between Brady and the NFL investigation was non-existent.
Given this lack of cooperation, I think it is the view of the NFL that neither the Patriots nor Brady deserve the most generous reading of the facts. Thru arrogance or malfeasance or both, I think the Patriots have managed to “inflate” the level of the penalties that they are now facing.
In the context of the other exculpatory evidence (including the other back and forth banter between McNally and Jastremski), I also find the suggestion that the moniker “Deflator” means something other than deflating the footballs to be rather stretched logic.
• Posted Jul 6, 2015 at 3:16 PM | Permalink
“But as we all know, but some may be less willing to acknowledge, the amount of cooperation between the New England Patriots and the league was very limited (only a single interview with McNally was permitted, and no followup interviews were allowed) and the contractually-obligated cooperation between Brady and the NFL investigation was non-existent.”
The League interviewed McNally 3 times before Wells was retained (all 3 times without any pats reps with him). the Patriots agreed to facilitate the fourth interview with Wells with an explicit understanding reached with the Wells investigators: barring unanticipated circumstances, individuals would only be interviewed by the Wells investigators one time. Wells had all the texts and phone records prior to his interview, brought 4 lawyers, and interviewed McNally for 7 hours. Brady also answered a full day of questions. The patriots gave the League and Wells access to everyone and everything they asked for, within the guidelines of their agreed upon interview rules.
You can read the details here: http://wellsreportcontext.com/#patriotscooperation
You can also look at the agreement and the dispute over the request for the 5th McNally interview here: https://wellsreportcontext.files.wordpress.com/2015/05/nep_psi_excerpts-from-emails-5_15_15.pdf
http://wellsreportcontext.com/#mcnallyfive
After you review all the details, do you still think the amount of cooperation between the Patriots and the league was very limited?
• Carrick
Posted Jul 6, 2015 at 10:38 PM | Permalink
davem1964, as I mentioned to Steve, as stated in the Well’s report, there was in fact a single interview between the NFL investigative team and McNally. The Patriots knew this and IMO were being very sleazy in how they characterized it. It would have been only a second interview, and the first after all of the physical evidence had been gathered and examined (and they were willing to go to McNally’s home town for the interview).
What’s really odd here is the refusal for the interview doesn’t seem to have come from the Patriots main office, but from their legal counsel. In retrospect it was probably poor legal advice as (I think) more of the punitive action occurred due to a perceived lack of cooperation than any actual concern over malfeasance on the Patriots part.
• MikeN
Posted Jul 7, 2015 at 10:08 AM | Permalink
I agree the stiff penalty on both the Patriots and Brady is for lack of cooperation.
Note however, that the NFL did have all the evidence they needed before the first interview. They just didn’t notice the deflator text because it was so far in the past.
• chuckrr
Posted Jul 6, 2015 at 3:26 PM | Permalink
McNally was interviewed four times, and Mr. Jastremski had been interviewed twice. Here’s another interesting fact….McNally ,who took the balls into the bathroom was not a full time employee of the Patriots. He worked on game days only. They put a lot of trust in someone that was a part time employee.
• Steve McIntyre
Posted Jul 6, 2015 at 3:28 PM | Permalink
Carrick, I think that some of your comments are at cross purposes here. The issue is not whether the PAtriot balls “were more deflated than the Colts balls”, but whether McNally tampered with the balls after they were approved by the referees according to a plan initiated or approved by Brady. It is evident that PAtriot balls were “more deflated” than Colt balls at half-time: that is not in dispute.
You say: “It is my personal speculate that the Patriots did not intend to under-inflate the balls, rather just to reduce them to the lower end of the permissible range.” I agree. But surely no one can object to that.
You say “the assumption of exponential warming is unlikely to be correct here” and somewhat snipe at me on these grounds, but it was the Wells Report that employed negative exponential transients to arrive at their conclusions of tampering. Even though they didn’t label their transients as negative exponentials, they clearly were.
You say: “I do not think the physics data (including metadata such as how the balls were handled before being measured) are good enough to say “this hypothesis is plausible but that one is not””. Fair enough. But Exponent made strong technical findings against the Patriots – finding that, in my opinion, could not be properly made on the facts available to them. If you feel that the technical data do not support findings either way, then it seems to me that you might have criticized the Wells Report more directly in that respect.
You say “I still think it the most parsimonious explanation based on all of the evidence is that the Patriots balls were simply inflated to a lower initial pressure, and that this was done deliberately by the Patriots staff, more likely than not in collusion with Tom Brady.” Let me suggest another possibility, based on an interesting reader comment. The Patriots definitely rubbed the footballs vigorously prior to measurement – very likely to get a slight edge at pregame measurement. Doesn’t seem to me like that infringes any rules, though it would definitely be working the system. According to Exponent, the footballs would have cooled to equilibrium prior to measurement and be under-inflated by a few tenths of a psi by the time of measurement. Footballs tendered at 12.2 psi or so ought to have been noticed by the referee. However, unbeknowst to everyone, one of referee Anderson’s gauges was off by 0.38 psi and he didn’t notice the under-inflation. Because he had two gauges and the other gauge, used for Colt balls, was unbiased – the other measurements were uneventful. But there’s a big legal difference between tendering rubbed balls to the referee in the hopes of gaming the system and tampering with the balls in the washroom – and your comments, as written, do not distinguish as clearly as they might.
An important physical point that, in my opinion, survives any dispute about the precise shape of the transients is ruling out “high” and even “moderate” deflation scenarios. I think that this is firmly supported on the available data; (2) that, if people had clearly understood that the physical data ruled out high and moderate deflation scenarios, they would have taken a different line on the entire case; and (3) that Exponent, as data analysts, ought to have informed their clients that high and moderate deflation were ruled out. Whether such failure was due to ignorance, negligence or prosecutorialness doesn’t matter, they ought to have put this information on the table.
with respect, it is the Wells Report that introduced negative exponential pressure transients. Merely because they did not label them as such doesn’t mean that they didn’t use them.
In my original comment, I observed that, in my opinion, the phenomenon of evaporative cooling ought to have been of concern and squarely addressed by the Wells Report and I criticized them for not doing so.
If evaporative cooling has more impact than allowed for in the negative exponential transients of the Wells Report, then it is to the advantage of the Patriots.
The Patriots also give a different account of their cooperation with the NFL. As I recall, they said that McNally was interviewed on four different occasions.
• MikeN
Posted Jul 6, 2015 at 5:11 PM | Permalink
It was not just prior but many hours before. The rubbing is to change the feel of the football, basically having its surface match the surface of the receivers’ gloves. It is not clear if the inflation to 12.5 happened immediately after each ball is treated, or all the balls at once.
Steve: I understand that the effect had almost certainly worn off. My point was quite different. One reader observed that the Patriot’s own measurements of pressure were made after rubbing, rather than before rubbing. I think that the evidence indicates that the original Patriot measurements were much closer to the time of rubbing relative to the later referee measurements. In this scenario, had both of referee Anderson’s gauges been properly calibrated, he would have picked up the under-inflation of Patriot balls at 71 deg F when they had further cooled. The whole thing seems like a SNAFU in its original meaning.
• MikeN
Posted Jul 6, 2015 at 6:35 PM | Permalink
I think that was my comment, but I misstated the effect.
The footballs were inflated to 12.5 after the rubbing. Exponent lists it as a .7 PSI difference but the time factor of the inflation could do exactly what you said.
However, the temperature of the place where the Patriots did the rubbing and inflated is not known.
• Carrick
Posted Jul 6, 2015 at 9:55 PM | Permalink
Steve McIntyre:
You say: “It is my personal speculate that the Patriots did not intend to under-inflate the balls, rather just to reduce them to the lower end of the permissible range.” I agree. But surely no one can object to that.
The problem is the rules as currently in place (which I think are flawed) do not permit the Patriots to “adjust” the ball pressure, after the Referee has taken possession of them. So I would imagine the Patriots would object to my statement at the least.
To be clear I think the penalty was too high (virtually anything more than a wrist slap was too high, and I still think the severity of the penalty has more to do with the perceived lack of cooperation on the part of the Patriots).
I also think the rule should be amended to permit what I presume Brady and “co-conspirators” to have been doing: I think allowing the teams to “tweak” the football pressure within the allowed range of values is in line with separating out the better coached teams from the poorer ones.
You say “the assumption of exponential warming is unlikely to be correct here” and somewhat snipe at me on these grounds, but it was the Wells Report that employed negative exponential transients to arrive at their conclusions of tampering. Even though they didn’t label their transients as negative exponentials, they clearly were.
I don’t question they employed exponentials here, but it is unlikely that this assumption is correct. I think you can use this assumption to demonstrate plausibility, but because the assumption is flawed, you can’t use it to rank hypotheses into more probable or less probable: The errors in the model probably aren’t quantifiable in this case.
The problem is that the solution of the heat equation in general admits to a series of exponentials with different coefficients. After a sufficiently long time, if the system remains undisturbed, you’ll end up with only the slowest decaying exponential remaining with a significant amplitude.
In practice, where the footballs are getting handled over time (so the modes are being continually “remixed”) and where (important) you have significant convective heat exchange (both nonlinearity, and much faster rate of heat transfer than predicted by simple heat conduction of air), I’d expect to see significant violations of the exponential rate of warming (the actual rate of warming will be faster, but not necessarily smooth).
Another effect that people who don’t do temperature measurements may not realize is that on a cold day, you end up with a strong negative vertical temperature gradient (the top of the building loses heat while more heat gets pumped into the bottom). This also leads to a faster rate of warming than you might measure on another, more temperate date. It’s worth mentioning here because if you want to replicate the real conditions of the original warming footballs, you have to get the room thermodynamics correct too. It’s not just an isolated football, it’s really a stack of footballs in or out of a bag, sitting in a room with or without air blowing on it from a heater.
Real-world thermodynamics is surprisingly more complicated than people whose entire experience in thermodynamics was with a well-stirred dewar might assume.
But Exponent made strong technical findings against the Patriots – finding that, in my opinion, could not be properly made on the facts available to them. If you feel that the technical data do not support findings either way, then it seems to me that you might have criticized the Wells Report more directly in that respect.
I think the case is much weaker than Exponent suggested (or at least that I perceive him to suggest, so in that sense I agree with what you have been doing. I think the strength of the Wells case is on the other evidence.
An important physical point that, in my opinion, survives any dispute about the precise shape of the transients is ruling out “high” and even “moderate” deflation scenarios. I think that this is firmly supported on the available data; (2) that, if people had clearly understood that the physical data ruled out high and moderate deflation scenarios, they would have taken a different line on the entire case; and (3) that Exponent, as data analysts, ought to have informed their clients that high and moderate deflation were ruled out. Whether such failure was due to ignorance, negligence or prosecutorialness doesn’t matter, they ought to have put this information on the table.
I’m not convinced the physics data really rule out high and moderate deflation scenarios. I think it’s plausible that if we did the physics correctly, we might be lead to the conclusion that the low deflation scenarios were ruled out (modulo the confounding factors that I think you correctly highlighted).
My gut feeling is the real rate of warming (combined effects of ball handling mixing up the modes + convective heat exchange) is likely faster than admitted to by Exponent’s and/or your modeling. Generally the warming rate you get when things are “stirred up” is faster than you’d get just from the slowest decaying mode. This is because heat energy gets transferred to faster decaying modes which in general leads to an increase in rate of heat energy transfer with the environment. Almost no likely scenario that I can think of would lead to a significantly slower rate of warming, except this one perhaps:
If evaporative cooling has more impact than allowed for in the negative exponential transients of the Wells Report, then it is to the advantage of the Patriots.
It’s something definitely worth looking at, but my guess without plugging numbers in, for a football that has internal convection and resides in a room with significant convective air motion, the relative contribution of evaporative cooling will be pretty minor compared to heat energy transfer from the exterior into the interior of the football.
The Patriots also give a different account of their cooperation with the NFL. As I recall, they said that McNally was interviewed on four different occasions.
I think the Patriots were a bit sleazy in this respect. As they well know, McNally was interviewed only once by the NFL investigative team. After discrepancies were found in McNally’s testimony, the Patriots refused to allow a follow up interview. From the Well’s report:
We believe the failure by the Patriots and its counsel to produce McNally for the requested follow-up interview violated the club‟s obligations to cooperate with the investigation under the Policy on Integrity of the Game & Enforcement of League Rules and was inconsistent with public statements made by the Patriots pledging full cooperation with the investigation.
Really (in my opinion) the Patriots and Brady are getting penalized for failing to fully cooperate with the investigation more so than for anything they did. Nobody questions that the Patriots would have won this game regardless. In fact, reducing the pressure (if it happened) probably had virtually no affect on the game whatsoever.
• Steve McIntyre
Posted Jul 6, 2015 at 10:50 PM | Permalink
Carrick, I will readily acknowledge that many people, including yourself, have much better knowledge of heat transfer issues than myself. In writing these posts, I’ve mainly focused on whether the Wells Report proved their point on their physics rather than trying to propose an alternative better physics. In that respect, I believe that my observation that higher deflation is ruled out is valid given the assumptions and evidence of the Wells Report, though, as you point out, this may not be valid if your surmise about faster warming rates is correct. However, that wasn’t what they presented.
My wondering about evaporative cooling is sincere and I would be interested in the opinion of a more knowledgeable observer. In 2006 experiments on leather and synthetic basketballs sponsored by Mark Cuban, the specialists observed that leather basketballs very quickly absorbed 70 g of moisture in far less trying conditions than Foxboro. The surface area of a football is about half that of a basketball, so, in the absence of proper analysis by Exponent, it seems plausible that leather NFL footballs in Game Day use would have absorbed at least 35 g of moisture. I have no experience in this calculation, but based on a little research, it seems to be that the 35 g of moisture, if evaporated, is enough to be relevant in a case where the dispute hinges on a couple of deg F of ball temperature. I would certainly appreciate an opinion from someone experienced in such calculations.
my guess without plugging numbers in, for a football that has internal convection and resides in a room with significant convective air motion, the relative contribution of evaporative cooling will be pretty minor compared to heat energy transfer from the exterior into the interior of the football.
• Carrick
Posted Jul 7, 2015 at 12:38 AM | Permalink
Steve McIntyre, if you had as much as 35 mg of water in the surface, the heat capacity of the water would clearly dominate the total heat capacity of the football. There is roughly 10-gm of air, but the heat capacitance of air at constant volume is only 0.7 kJ/kg/K compared to water with 4 kJ/kg/K.
Anyway, it’s a complicated problem because the water is in the skin. Probably this could accurately calculate this (without trying to formulate it, this seems like a graduate student level problem), but it’d be a lot simpler to measure the effect of moisture in the skin on the rate of warming.
• mpainter
Posted Jul 7, 2015 at 3:59 AM | Permalink
Carrick,
The dominant fact of the official’s room at half-time is the low humidity, given at 20%. Exponent did not address the possibility of evaporative cooling of the wet balls and this failure is clearly a defect of their study. IMO the wet balls would not have started warming until they had dried and this means that the gauge readings at half-time would fall below Exponent’s transient curves.
Since the first half ended with an eleven play drive by The Patriots, their balls would have been wet, and indeed, the Wells Report confirms this.
• Carrick
Posted Jul 7, 2015 at 9:30 AM | Permalink
mpainter:
The dominant fact of the official’s room at half-time is the low humidity, given at 20%. Exponent did not address the possibility of evaporative cooling of the wet balls and this failure is clearly a defect of their study. IMO the wet balls would not have started warming until they had dried and this means that the gauge readings at half-time would fall below Exponent’s transient curves
That Exponent did not address whether the balls were wet, does not appear to be true. From the Wells report:
According to Exponent, the environmental conditions with the most significant impact on the pressure measurements recorded at halftime were the temperature in the Officials Locker Room when the game balls were tested prior to the game and at halftime, the temperature on the field during the first half of the game, the amount of time elapsed between when the game balls were brought back to the Officials Locker Room at halftime and when they were tested, and whether the game balls were wet or dry when they were tested at halftime.
In any case, the footballs are kept in a ball bag until used. Since they don’t generally go through all 13 balls in a half, some of them would likely have been relatively dry.
IMO the wet balls would not have started warming until they had dried and this means that the gauge readings at half-time would fall below Exponent’s transient curves.
It’s certainly not the case that the wet surface prevents heat exchange between the external environment and the interior. I’d expect the skin to warm over time, even with evaporative cooling, just at a slower rate. As I suggested to Steve, this is much easier to do using a real football, rather than trying to model. (And certainly any model would need to be validated with data.)
• Steve McIntyre
Posted Jul 7, 2015 at 11:30 AM | Permalink
Carrick, I think that you’re somewhat at cross purposes with the comment. The issue with Exponent and wetness is not that they didn’t recognize it, but that they didn’t explain why wetness would impact pressure through reference to physics principles or fully analyse its effects.
In some early commentary, it was speculated that the volume of wet footballs could change due to elasticity. Exponent analysed volume changes and showed – convincingly in my opinion – that volume was not impacted by wetness. However, they didn’t follow up on this observation. Because wet footballs are still subject to the Ideal Gas Law, my understanding is that, if wet footballs have lower pressures, then the ball temperatures must be less than the ball temperatures of dry footballs. And thus presumably evidence of evaporative cooling.
The negative exponentials shown in Exponent’s figures were based on their simulations, rather than theory. In that respect, it seems to me that you’re being inconsistent in, one on the one hand, pooh-poohing analysis based on the negative exponentials derived from Exponent’s experiments, on the grounds that the actual warming might be much faster than shown in the various transients in my discussions (which are based on Exponent data), while on the other hand saying that the behavior is best determined by simulations.
In this respect, if Exponent released a compendium of simulation results, it would resolve some important questions. In particular, their GameDay simulations – which seem instructive to me – are reported only in the aggregate, so we don’t see what inter-ball variability existed under Game Day simulation conditions, which are purported to simulate varying wetness. Release of this data would help.
• mpainter
Posted Jul 7, 2015 at 10:23 AM | Permalink
Carrick,
Exponent certainly did _not_ address the issue of evaporative cooling and this has been commented on by sources in the media as well as Steve here.
The low humidity of the official’s room is the prime consideration and the rate of evaporation should have been high enough to affect temperature of the air inside the balls.
It would have been a simple matter to devise an experiment meant to recreate the circumstances in order to determine if wet balls cooled at half-time,or if they simply failed to warm until they had dried.
Exponent failed to make this important determination, and indeed did not consider the possibility.
The question of the degree of wetness, and how many of the balls were wet and how many were dry will probably never be determined.
Regarding the Patriots’ eleven play before half-time, it was raining and I assume that after each play the Patriots’ bench sent out a (relatively) dry ball. In this case, eleven balls would have been wetted just prior to halftime.
• mpainter
Posted Jul 7, 2015 at 10:38 AM | Permalink
Carrick,
Whether or not the wet balls cooled is best determined by a demonstration under identical circumstances.
Evaporative cooling can quite reasonably be expected in the circumstances. Whether any particular ball cooled a bare fraction of a degree or a greater amount cannot be determined.
As I have said before, the failure to measure corresponding ball temperatures at the half-time fiasco renders the whole study inconclusive.
• Steve McIntyre
Posted Jul 7, 2015 at 11:41 AM | Permalink
Whether or not the wet balls cooled is best determined by a demonstration under identical circumstances.
According to my calculations, the only way to explain a ~0.3 psi pressure drop for wet balls is through a decrease in ball temperature of about 5.75 deg F.
According to the Ideal Gas Law (and calculations in Wells Report), the half-time pressure at 48 deg F of Colt balls initialized at 13 psi at 71 deg F is 11.8 psi. The corresponding pressure of wet balls is said to be ~0.3 psi less (though some diagrams show a greater difference).
Since the wet balls were initialized under the same conditions, they have the same number of moles of air and, according to Exponent, there was no volume change for wet balls. The corresponding ball temperature for balls at a pressure of 11.5 psi is 42.25 deg F. It’s regrettable that Exponent made no attempt to explain.
I’ve seen some commentary which argued that evaporative cooling would not be a factor while the balls were still outside in saturated conditions, but this seems to be contradicted by Exponent’s Figure 21 which shows lower pressure in wet balls even in their simulated outside conditions.
• mpainter'm
Posted Jul 7, 2015 at 2:24 PM | Permalink
Steve:
“The only way to explain a .3 psi pressure drop for wet balls is through a decrease in ball temperature of about 5.75 deg F.”
####
I agree, via evaporative cooling.
Exponent’s claim that wetness does not affect volume has a dubious flavor.
I have not read Exponent’s study, but a ball that evaoratively cools will undergo a temperature reduction and hence a pressure reduction. I do not understand why they performed volume analysis.
Steve: one of the earlier comments on the controversy by Headsmart labs speculated that wet footballs might have experienced slight volume increases, thus a contributing factor to pressure decrease. This was immediately questioned, with evaporative cooling posed as an alternative. Exponent showed no volume changes – I think that it was prudent to rule this out and do not criticize them on this point, though I think that they should have discussed why wet footballs had lower pressure.
• Steve McIntyre
Posted Jul 7, 2015 at 3:22 PM | Permalink
As I noted in my original writeup, there are numerous online demonstrations of evaporative cooling in other contexts, e.g.
https://www.youtube.com/watch?v=cwyDsHG-ymQ. For someone, like myself, who’d not previously thought about the phenomenon, I though that these demonstrations were pretty interesting.
Is it possible that evaporative cooling might have intensified when the balls came into the dry air?
• mpainter'm
Posted Jul 7, 2015 at 4:47 PM | Permalink
Steve,
The NOAA handy-dandy dewpoint calculator gives, at 73°F and relative humidity of 20%, a dewpoint of 29.58 °F.
This means that evaporative cooling in the official’s room at half-time could drop the ball temperature by 40°F, theoretically. The porous leather holding water is analogous to the cotton gauze of the demonstration. What is lacking is the advection of the blow dryer.
The low humidity is the key. It should be possible to get some theoretical cooling on a case basis, meaning given, say, 10 grams of water held in the leather shell of the ball. The heat of evaporation for 10 g of water is over 20,000 joules.
This heat requirement is axiomatic and will pull as much heat from the air contained inside the football according to the conduction properties of the ball materials. All sorts of considerations.
• Steve McIntyre
Posted Jul 7, 2015 at 7:52 PM | Permalink
Here is the source for absorption of moisture in 2006 by leather NBA balls: http://blogmaverick.com/2006/10/27/nba-balls/).
• mpainter'm
Posted Jul 7, 2015 at 5:09 PM | Permalink
It seems plain that had Exponent delved into the possibilities of evaporative cooling of the balls at half-time, they would have torn a ragged hole in the prosecution brief. Hard to keep clients when you go around tearing holes in their briefs.
• mpainter'm
Posted Jul 7, 2015 at 7:54 PM | Permalink
Correction:
The balls arrive at a temperature of 48°F. Theoretical evaporative cooling would be to 32°F.
• Carrick
Posted Jul 8, 2015 at 2:50 AM | Permalink
I think you’ll find the real world effects of evaporative cooling to be considerably more complicated than seems to be imagined here.
Air currents play a big role in promoting water vapor exchange, hence “wind chill factor”.
So when the footballs are inside of the football bag, they are going to be protected from air currents, and the influence of any evaporative cooling is going to be squelched. The position of the top of the football bag is important here
Secondly, when a football gets removed from the bag and placed on the table, the bottom will be in contact with the table (which is radiative heated and can be treated as a heat source).
The top of the football will probably begin cooling due to evaporation induced by air motion above the football, but that sets up a negative temperature gradient within the football, which in turn promotes air convection internal to the football.
I think you’d end up with a greater rate of heat energy exchange because of the air motion interior to the football with evaporative cooling that you would without it.
Also you guys seem to be equating the skin temperature of the football with the interior temperature. When there aren’t temperature gradients, this would be the case I think. Not so, when you have a football sitting on a warm surface, like the table.
• Steve McIntyre
Posted Jul 8, 2015 at 10:27 AM | Permalink
Carrick, I think that you are being very unfair in criticizing us for pondering these issues, when the \$5 million Wells Report, which was charged with investigating the matter, failed to do so. In approaching matters from an “audit” perspective – really, nothing more than detailed peer review – I am generally wary of attempting my own solution to thorny problems, as opposed to assessing whether the specialists have proven their point. In this respect, I think that you’ve rather pulled your punches in commenting on whether Exponent had established their technical finding. My primary point has been whether the Wells Report proved their strong claim that the Patriot pressures could not be explained under physical principles, rather than saying that evaporative cooling was the magic elixir. I raised evaporative cooling as an issue because it seemed interesting and relevant and because it was not foreclosed in the Wells Report – which ought to have done so – rather than promoting it as THE answer. Again, from a statistical and data analysis perspective, I believe that the most troubling aspect of Exponent’s technical report was their false claim to knowledge on points that remain indeterminate. (Actually, I started my own analysis with their statistical “model” which I haven’t written about, but it is appallingly bad data analysis. It’s so bad as to be uninteresting in respect to determining what happened. But I should really write it up purely as an exercise in bad statistics – especially since it touches on random effects models, a technique that I’ve written about on many occasions and about which I’m knowledgeable.)
)
• Carrick
Posted Jul 8, 2015 at 10:31 PM | Permalink
Steve McIntyre:
Carrick, I think that you are being very unfair in criticizing us for pondering these issues, when the \$5 million Wells Report, which was charged with investigating the matter, failed to do so.
Sorry but this seems like a diversion to me. If there are problems with your criticisms of the Exponent report, I think it’s fair game for me to bring this up.
You accuse me of being too lenient on Exponent. Before I address that, as far as I can tell, the only thing that Exponent’s report was used for was to establish plausibility to the claim that McNally intentionally deflated the Patriots footballs.
I think this was successfully done, your arguments to the contrary notwithstanding. I will remind you of this text from the Well’s report:
In reaching the conclusions set forth in this Report, we are mindful that the analyses performed by our scientific consultants necessarily rely on reasoned assumptions and that varying the applicable assumptions can have a material impact on the ultimate conclusions. We therefore have been careful not to give undue weight to the experimental results and have instead relied on the totality of the evidence developed during the investigation. Even putting aside the experimental results, we believe that our conclusions are supported by the evidence in its entirety.
Now I think this is the appropriate weight to put on this analysis, because in my opinion there are just too many problems with the measurement protocol itself to allow a unique interpretation of the events that lead up to that particular collection of measurements.
I felt this way before reading the narrative text and appendices containing Exponents report. The information that has been provided to the public in the Wells report on the Exponent study was surprising lean, and not consistent with scientific standards for peer review. But I’m hardly surprised to see what appears to be a slopped together mess (from the few details that have been publicly released).
The purpose of my comments here were to inject some realism into the discussion of … physics. Hopefully nobody will object to that.
You said at the start of your post that:
This is a sensible observation, but raises the question of whether and how one could use the available statistical information to exclude tampering. This is analysis that ought to have been done in the Wells Report. I’ve done the analysis in this post and the results are sharper than I’d anticipated.
The trouble I had here is this problem is in the realm of physics, not statistics. You can’t simply choose a “toy” physics model (without testing it) and use that to decide which of the set of available hypotheses is more likely. As I pointed out, exponential relaxation is almost certainly the wrong physical model here.
You argued that “Exponent used it too”. Yes they did. But they used it to demonstrate consistency of the data with the hypothesis that the footballs had been tampered with. The use of a “toy” model (one that is physical realizable) is a reasonable thing to do, if all you are trying to do is establish that it is plausible, based upon the data, that the footballs had been tampered with.
It would have been an end of the case if they couldn’t find a physically plausible model that explained the data in a way that pointed to tampering. Putting more weight on Exponent’s work beyond that is in my opinion a waste of time.
I will note that the Wells report contains this gem:
Based on the testing and analysis, however, Exponent concluded that, within the range of likely game conditions and circumstances studied, they could identify no set of credible environmental or physical factors that completely accounts for the Patriots halftime measurements or for the additional loss in air pressure exhibited by the Patriots game balls, as compared to the loss in air pressure exhibited by the Colts game balls. Dr. Marlow agreed with this and all of Exponent‟s conclusions.
I think you have demonstrated that at least one credible scenario exists.
But I think you are very far from showing that the physical data + realistic modeling precludes tampering with the football. (Again, it’s my opinion that it is impossible due to the general weaknesses in the measurement protocol to reach an definitive conclusions.)
One thing you probably want to consider is that the Patriots footballs were treated differently than the Indy footballs were treated. This is a serious confound in trying to invoke evaporative cooling, as what I have read suggests that the Indy footballs were coated with less oil making them more susceptible to evaporative cooling than the New England footballs.
Anyway, I don’t take Exponents report very seriously because Wells (eventually) did not put much weight in it. You seem to want to make much bolder claims. Bolder claims deserve more close scrutiny.
• Steve McIntyre
Posted Jul 8, 2015 at 11:16 PM | Permalink
Carrick, c’mon, you’re being unfair here. You say:
I don’t take Exponents report very seriously because Wells (eventually) did not put much weight in it. You seem to want to make much bolder claims. Bolder claims deserve more close scrutiny.
I entirely disagree with your statement that the Wells Report “did not put much weight in it [Exponent’s report]” or that it only used Exponent’s report to “establish plausibility to the claim that McNally intentionally deflated the Patriots footballs”. The Wells Report quoted and relied on Exponent’s findings for much more than that: Exponent’s assertion that (1) Patriot ball pressures could not be accounted for by known physics; and (2) there was an unexplainable discrepancy between Patriot and Colt pressures, fundamentally colored and, in my opinion, poisoned the subsequent perception of the lawyers. You quote what you describe as a “gem” – an incorrect finding that was central to my original writeup. If Exponent had stated that it was possible that physical factors could have accounted for observations – as it ought to have done, events surely would have unfolded differently. In particular, I don’t see how anyone can arrive at a fair disposition when the record is contaminated by a technical report as faulty as Exponent’s.
You argue that I’ve over-reached in arguing that the data precludes “high” or “moderate” deflation scenarios on the grounds, as I understand it, that the data is simply too poor to permit any conclusions. Nearly all of my articles on this have been been focused on criticisms of the Wells Report rather than trying to draw my own conclusions from the data. I’ve repeatedly criticized the inadequacy of both NFL protocols and Exponent’s analysis.
In this particular blog post, I have attempted to wring something out of the data and while you disparage the effort as being unwarranted by the data, notwithstanding your criticism, I think that my deductions from the data are better supported and argued than Exponent’s deductions from the same data.
• Carrick
Posted Jul 9, 2015 at 9:53 AM | Permalink
mpainter:
You can only make such a statement by strangely ignoring the potential for evaporative cooling.You have ridiculed our discussion of this and now You say “Crazy talk” and this reveals something about yourself: a ball that cools cannot warm and you miss that.
I’m not ignoring the potential for evaporative cooling. I’m saying you are simply wrong that evaporative cooling prevents an object from warming until it’s completely dry. Are you thinking of a pot boiling or something?
The physics here is nothing like that. Evaporative cooling just means you have a mechanism for heat loss. If you have more heat energy being transferred between the environment than is being lost by evaporative cooling, the football will warm, whether it is wet or dry. In general, because the actual amount of water absorbed into the football is modest, you should always see warming, whether the ball is wet or dry, and whether the air is humid or not.
• Carrick
Posted Jul 9, 2015 at 4:56 PM | Permalink
Steve McIntyre:
I entirely disagree with your statement that the Wells Report “did not put much weight in it [Exponent’s report]” or that it only used Exponent’s report to “establish plausibility to the claim that McNally intentionally deflated the Patriots footballs”.
You’re entitled to disagree of course, but to do so you have to selectively ignore key parts of the Wells report, such as where it said:
In reaching the conclusions set forth in this Report, we are mindful that the analyses performed by our scientific consultants necessarily rely on reasoned assumptions and that varying the applicable assumptions can have a material impact on the ultimate conclusions. We therefore have been careful not to give undue weight to the experimental results and have instead relied on the totality of the evidence developed during the investigation. Even putting aside the experimental results, we believe that our conclusions are supported by the evidence in its entirety.
I read this in the plain English sense that the conclusions of the report did not hinge on the results from Exponent.
• mpainter'm
Posted Jul 8, 2015 at 11:21 AM | Permalink
Steve,
The basketball illustration well shows the absorptive capacity of leather.
However, the report gives that the balls were used with a somewhat worn surface and this would have increased the initial rate of absorption, it seems.
New NFL football’s would have a lower rate of absorption, I would guess.
But then there is the “rubbing” complication, which I have read is meant to remove the new ball finish, as a preference of some quarterbacks.
I know nothing about all of this, but it would seem that the degree of evaporative cooling depends on the amount of water held in the leather shell of the ball,and this depends on the absorbent properties of the Patriot balls.
If I were to simulate the problem, I would do it on a case basis, meaning, 2 grams of absorbed water, 4 grams, 10 grams, etc. and plot a curve of cooling versus wetness, with all pertinent conditions reproduced.
Carrick makes a good point about the balls being kept in the bag at half-time, but were they? This is another question for I have seen balls carried in a net bag.
• Steve McIntyre
Posted Jul 8, 2015 at 12:19 PM | Permalink
You say: “New NFL football’s would have a lower rate of absorption, I would guess”. Not sure about that. The purpose of the rubbing protocol was to remove the protective surface and make the balls easier to grip. Does anyone overtly rough up basketballs in the same way? I don’t think so. Obviously, play on a court would rough the surface up, but it seems plausible to me that aggressive roughing up could accomplish the same or more. I think that one has to assume just as much absorption per surface for footballs as basketballs (if not more). But yes, I agree that different simulations would settle the matter. Too bad the NFL didn’t get the analysis from Exponent.
I agree that balls in the bag at halftime is relevant and that this was something relevant that Exponent ought to have simulated. I recall seeing somewhere that Colt balls were both in a carry-bag, but further protected by a green plastic garbage bag.
My speculation on the impact of a (say, canvas) bag is that it would sort-of act like an inefficient layer of clothing or inefficient thermos- keeping something warm warm and something cool cool by lowering heat exchange. In other words, it would somewhat delay the warming of the balls at half-time. Since Exponent’s argument rests largely on the argument that pressure gain (warming) commenced immediately on entry to the officials’ room and that the Patriot balls ought to have had more pressure gain by the time that they were measured than was observed, any delay of warming due to canvas bag protection works against Exponent and in favor of the Patriots.
• mpainter'm
Posted Jul 8, 2015 at 2:03 PM | Permalink
Steve,
I agree that the bag would have prevented warming. If the ball bag was enclosed in a polyethylene garbage bag, that slams the door shut on gas exchange and leaves conduction as the only possibility and that would have been of an insignificant effect, if any.
But were the balls left in the bag or had some provision been made for the convenience of the referee, such as a wire bin to hold the balls while the referee did his pre game pressure checks?
All these questions and more. If the Colt’s balls were left in their bags, these would not have warmed, either.
• Steve McIntyre
Posted Jul 8, 2015 at 2:17 PM | Permalink
Here is a picture of what appears to be the Patriot ball bag. It looks like a vinyl duffle bag of some sort.
• mpainter'm
Posted Jul 8, 2015 at 2:15 PM | Permalink
Also,
I of course disagree with Exponent’s conclusion that wet balls would warm immediately. In this respect, see my comment on the previous post, (Bodge or Botch?) at July 6, 3:45 pm, the last comment.
In this comment, I review the fig 21 of the Wells Report, which seems to me as egregious and even incomprehensible.
• Carrick
Posted Jul 8, 2015 at 11:44 PM | Permalink
mpainter, I’m going to comment on your comment from the other thread here:
Note that the wet ball transient at halftime shows wet ball re-pressurization at the same rate as the dry balls for the first few minutes. This is impossible, if I’m not mistaken. I do not see how wet balls can begin to warm/re-pressurize until they have become dry balls.
Um, why do you think wet footballs can’t warm until they are dry? Crazy talk.
I’d be interested in hearing this spelled out, so if for no other reason, I can hopefully dispel you of this notion: I am certain you are wrong here and I believe the data are likely correct. (Note I’m not claiming though that these are the same waring curves the footballs tested in the locker room would have experienced.)
By the way, thanks for pointing me to this figure. I admit (as I have above) that I haven’t to this point taken the Exponent report very seriously. It is treated as weak evidentiary value, and I wasn’t particularly impressed with the parts I did read.
But now it’s clear that they did at least compare wet versus dry balls (even if they didn’t directly attempt to model the evaporative cooling, something that I think would be truly an Herculean effort except in the most of controlled scenarios).
It’s also interesting to look at the heating/cooling curves, which are again as I predicted distinctly non-exponential in their shape. If you try to match up the initial linear behavior to what an exponential would predict, you end up with a huge error in the later transient response. If you try and match up the later transients portion, you get too steep of an initial warming/cooling.
Regarding the footballs in the carrying bag, I agree that the bag will affect how the footballs warm. It’s erroneous to claim they won’t warm though:
If the footballs are sitting on the floor of the locker room, they will still get heat energy exchanged. It would not behave the same as loose closing, because of the large thermal contact of the bag with the floor. However, I’d expect a big gradient in temperature. The footballs at the bottom would be nearer to the floor temperature than the ones at that top, at the point where they are pulled out of the bag.
How the ball is tested makes a big difference here too. If all of the footballs are laid on the table before being tested, and I favor this is as the most plausible scenario for what was actually done, I’d expect to see a much accelerated warming curve (for wet and dry balls) compared to the Exponent measurements. [Again this points to the weakness of the measurement protocol that this vital information apparently was not recorded.]
A couple of notes on the Exponent tests:
The Exponent cooling/warming tests were performed with the football placed on what appears to be a football tee. That has a much lower surface contact than the footballs on the table scenario.
As I pointed out, the room conditions really matter here too. It is stated that the Exponent measurements were performed in a temperature controlled chamber. This has very different thermodynamic properties than a heated room in cold weather. On a cold day in a heated room, the ceiling is typically cold compared to the floor. This leads to strong convective mixing. Higher convection would yield a more rapid equilibration of the footballs with the room temperature than was seen with the Exponent measurements.
As I said above, I thought it was more likely than not that, based on certain assumptions about how the measurement process was being done, you could eliminate as plausible any scenarios that didn’t involve some deliberate manipulation of football pressure by McNally.
However, I’m going to emphasize that one should not over sell such a result—there are too may uncontrolled aspects of the measurements which could tip the balance in unexpected directions for me to buy into any conclusion that eliminates alternative scenarios.
• Steve McIntyre
Posted Jul 8, 2015 at 11:51 PM | Permalink
Carrick, you say:
It’s also interesting to look at the heating/cooling curves, which are again as I predicted distinctly non-exponential in their shape. If you try to match up the initial linear behavior to what an exponential would predict, you end up with a huge error in the later transient response. If you try and match up the later transients portion, you get too steep of an initial warming/cooling.
I’ve digitized all of the half-time transients in all of the figures and they can be very closely fitted by negative exponentials or at most a biexponential, though the wet transient of Figure 21 requires a biexponential. It might be worthwhile for me to post up a technical post showing this. As I mentioned in an earlier post, the more troubling aspect of Exponent’s simulations is that the dry transients of Figure 27 (Logo) do not appear to me to be achievable with the stated initialization using the Logo gauge. They start out too high.
• Steve McIntyre
Posted Jul 9, 2015 at 2:16 PM | Permalink
Carrick, you say:
A couple of notes on the Exponent tests:
The Exponent cooling/warming tests were performed with the football placed on what appears to be a football tee. That has a much lower surface contact than the footballs on the table scenario.
This is inconsistent with Exponent’s description of their Game Day tests, which also involve the measurement of footballs using gauges. From the description, it sounds like they tried to replicate conditions of balls and rooms in a realistic way and did NOT carry out their “half-time” measurements with the balls on a tee. This does not mean that their “officials’ room” necessarily replicated all relevant aspects of the actual officials’ room, but it does mean that your proffered criticism is invalid.
• Carrick
Posted Jul 9, 2015 at 12:48 AM | Permalink
Steve, if you just visually inspect the curves, they are initially too flat and the knee point to sharp to be explained by a single exponential. But I’ve digitized Figure 17 just to confirm what my eyes were telling me (one exponential does not give a good match to the data).
I’d be careful about reading too much into successfully fitting to two exponentials, beyond that saying the behavior is not purely exponential. But unfortunately, without a physical model, using two exponential functions becomes just a curve fitting exercise.
One exponent is predicted from Newton’s law of heat, so there’s a physical basis for it. Multiple exponentials are predicted from the heat equation, but there you can calculate all of the constants. If the system is allowed to cool without interference and IF there is no thermal convection, you’ll end up with a single exponential after a long enough time.
I believe you’d eventually get a single exponential even when you have nonlinearity involved from convection and the football is left undisturbed. When the temperature difference gets small enough, the convective heating becomes insignificant. However, the shape of the curve in region we care about—the initial transients—I’d expect the exponential constants of the faster transient(s) to vary significantly based on the measurement environment.
In a room with HVAC, you have a constant air current and you can forget about finding simple exponential behavior. (The temperature in the room is not constant, and you’ll find that the football temperature quickly starts tracking the variation in temperature in the room.)
But my biggest worry about non-exponential behavior comes from concern about episodic events: Removing the footballs from the carrying bag, setting them on the table, etc. That’s why I think Figure 17 and follow ons are basically worthless: They tell us nothing useful about the actual temperature or pressure course of the footballs as they were handled in the real world.
Good enough again for plausibility hand-waving arguments, but not useful for much more.
As I mentioned in an earlier post, the more troubling aspect of Exponent’s simulations is that the dry transients of Figure 27 (Logo) do not appear to me to be achievable with the stated initialization using the Logo gauge.
I wouldn’t have used the word simulation here: They are physical measurements performed in a temperature controlled chamber.
I’d say don’t expect these cheap gauges to be super-relaible. I wouldn’t be surprised if there was some nonlinear hysteresis here, amongst other issues.
I’m not going to get too much into issues with measuring gauge pressure on a football (where the membrane is not completely pliable), but lets just say there are issues here. You might not have thought of it, but the pressure in rooms that have HVAC installed is larger than in rooms where it is not present.
• Carrick
Posted Jul 9, 2015 at 2:01 AM | Permalink
Also, to be clear, I’m focusing till now on the warming portion of the curves, since that is what is of importance for analyzing the NFL officials data. There does seem to be a fair amount of asymmetry between the warming and cooling portions, which is interesting.
• mpainter'm
Posted Jul 9, 2015 at 4:54 AM | Permalink
Carrick says
“Um, why do you think wet footballs can’t warm until they are dry. Crazy talk.”
####
You can only make such a statement by strangely ignoring the potential for evaporative cooling.You have ridiculed our discussion of this and now You say “Crazy talk” and this reveals something about yourself: a ball that cools cannot warm and you miss that.
You embrace the transients devised by Exponent in fig 21 of the Wells Report. This transient shows the wet balls still warming after half-time when they have been returned to the field. Pretty muddle-headed science, imo.
• Carrick
Posted Jul 9, 2015 at 9:54 AM | Permalink
mpainter, I replied on the wrong sub-thread, sorry.
See my comment here.
Briefly—you have the physics wrong here.
• Carrick
Posted Jul 9, 2015 at 4:49 PM | Permalink
Steve McIntyre:
This is inconsistent with Exponent’s description of their Game Day tests, which also involve the measurement of footballs using gauges. From the description, it sounds like they tried to replicate conditions of balls and rooms in a realistic way and did NOT carry out their “half-time” measurements with the balls on a tee. This does not mean that their “officials’ room” necessarily replicated all relevant aspects of the actual officials’ room, but it does mean that your proffered criticism is invalid.
I’m not sure exactly which part of the report you’re discussing, so to be clear I was referring to the measurement conditions for Figs 17 and 21. The experimental setup, such as it was, delves more on the simple process of setting the temperature of the environmental chamber than it does on the details of how the footballs were mounted in the chamber.
Here, a detailed explanation of the set up is lacking, but I see nothing in their discussion that is inconsistent with my assumption that the setup was similar to Figure 20, which immediately proceeds this discussion.
Regardless of intent, I think it is likely that the footballs were indeed on a tee or something similar to it. It is clear they were performed inside of an unspecified environmental chamber, so in this respect there is little semblance between the measured conditions and the environmental exposure of the footballs on the game day.
In other words, I find it to be exceedingly unlikely that the experimental conditions used for these measurements were an acceptable approximation of the game day conditions to which the footballs were subjected.
9. EdeF
Posted Jul 5, 2015 at 10:36 PM | Permalink
Right before the start of the American Football Conference Championship game, in front of 70,000 fans, I would not be surprised that someone’s urge before running out onto the field would be to visit the restroom. Now, taking a look at Fig 1 above, and setting the timing in reverse, if the footballs are inflated to 12.5 psi at 70 F in the locker room, it looks like within 15 minutes they are going to be deflated down into the 10-11 psi range for a good part of the first half, or second half. This is right in Brady’s sweet spot. Why deflate the balls anymore? What good would it do? I can see a ballboy checking the balls to make sure they are not 16 psi, but they are good at 12.5.
10. Posted Jul 6, 2015 at 1:20 PM | Permalink
There is added context for group (1) text “Talked to him last night. He actually brought you up and said you must have a lot of stress trying to get them done.” According to the Patriots response, http://wellsreportcontext.com/ : “After the conversation with Mr. Jastremski’s friend was explained by Mr. Jastremski, the investigators did not request the opportunity to interview the Mr. Jastremski friend to determine whether any such conversation had in fact happened. The Patriots tracked down Mr. Jastremski’s friend, who is a professional fraud investigator and whose livelihood depends on his honesty. They arranged for a telephone interview with the investigators in which the individual explained in great detail the timing (the night of the Jets game), place (Mr. Jastremski’s house) and content of the conversation (dealing with Mr. McNally’s sister, suffering some early onset memory loss, trying to sell the family game tickets). The investigators, rather than take further steps to check out this information, simply chose to disbelieve input that did not square with their conclusions.”
As others have noted, the Wells investigation and report seems more like a prosecutorial advocacy paper vs. an independent investigation searching for truth.
11. MikeN
Posted Jul 6, 2015 at 5:03 PM | Permalink
http://www.backpicks.com/2015/05/17/the-cognitive-and-statistical-biases-of-deflate-gate/
A Lack of context and predictable cognitive biases can make text messages appear as they aren’t, and make alternative explanations less believable than they really are
12. Posted Jul 6, 2015 at 7:53 PM | Permalink
MikeN – here’s the gloving process from pg. 50 of the wells report: “Jastremski told us that he set the pressure level to 12.6 psi after each ball was gloved and then placed the ball on a trunk in the equipment room for Brady to review.”
If you interpret that as: 1. glove ball, 2. set psi to 12.6, 3. place ball on trunk, 4. repeat 1-3 for next ball, and so on… which I think is 1 reasonable interpretation of that statement, then the pats balls were really around 12 or 12.1 when Anderson checked them pre-game.
This would be a major blow to the Wells report assertion that Anderson’s recollection of using the logo gauge to check pats balls was wrong because pats balls wouldn’t read 12.5 on the logo gauge since it reads .4 higher (and pats set their balls at 12.6).
If the pats balls were around 12.1, Now Anderson’s recollection of using the logo gauge makes sense: The logo gauge reads .4 higher, so he’d read pats balls on the logo gauge at 12.5 – which is how he recollects pre-game – logo gauge, 12.5 psi.
Isn’t this scenario (my scenario 1) more likely to have occured?: pats glove ball and set to 12.6, glove effect wears off and balls are at 12.1, Anderson uses logo gauge (which he recollects), logo gauge reads .4 higher, so balls read around 12.5 (which Anderson also recollects). Then at halftime, the logo gauge #’s fall within what’s expected, without tampering.
The scenario we have to believe to be true to show tampering is: pats set balls to 12.6 (no gloving effect). Anderson uses non-logo gauge (which he doesn’t recollect) and reads pats balls at 12.5. And then the half-time non-logo gauge readings show .4 psi tampering. So in this scenario, we believe McNally must have taken .4 psi out of each ball in the bathroom.
I believe my scenario 1 makes more sense given all the gauge and pre-game/halftime info in the wells report.
As far as some of the other indirect (non-science) evidence of tampering that previous posters have referred to:
1. “Deflator” – McNally and Jastremski called McNally the deflator 1 time (McNally referred to himself) – in the offseason between the 2013 and 2014 season. So to follow the logic that this shows McNally was deflating footballs after pre-game check means that he was doing this during the 2013 season and for all of the 2014 season. How does one explain the balls being over-inflated at 16 psi in the 2014 Jets game if he was deflating the balls below 12.5 psi after the pre-game check? This doesn’t make sense to me.
2. The text: “Talked to him last night. He actually brought you up and said you must have a lot of stress trying to get them done.” There is no record of Brady calling or texting “last night”. Jastremski explained this was about something and with somebody (a friend) completely different. the investigators did not request the opportunity to interview the friend to determine whether any such conversation had in fact happened. The Patriots tracked down the friend, who is a professional fraud investigator and whose livelihood depends on his honesty. They arranged for a telephone interview with the investigators in which the individual explained in great detail the timing (the night of the Jets game), place (Mr. Jastremski’s house) and content of the conversation (dealing with Mr. McNally’s sister, suffering some early onset memory loss, trying to sell the family game tickets). The investigators, rather than take further steps to check out this information, simply chose to disbelieve input that did not square with their conclusions.
3. Increased communication between Brady and Jastremski after the story first broke. If a story just broke where your team is accused of deflating psi by 2 lbs (which was false) and at the time no one even knew that they should naturally be 1 psi below what they started at… wouldn’t you talk with the person in charge of preparing the balls to try to understand how this could have happened?
4. McNally stopping to use the bathroom on the way to bringing the balls to the field. If one thinks my scenario 1 above makes more sense, then one reasonable explanation is he stopped to relieve himself on the way to the field before he worked outside for the next hour.
• MikeN
Posted Jul 7, 2015 at 9:15 AM | Permalink
The Wells Report gives the gloving an effect of .7psi. This could explain why a few of the footballs came in under 12.5 and had to be reinflated.
13. jddohio
Posted Jul 8, 2015 at 10:59 AM | Permalink
Steve: I commend you for thoroughly examining Exponent’s analysis and pointing out its flaws.
However, I quickly read the Wells report and there is strong circumstantial evidence of tampering with the balls separate and apart from any flaws in the testing that was done.
First, according to the officials, it was the typical practice to leave the balls in the officials locker room. They were surprised when McNally removed the balls. (p56). When asked why he did so after the game, McNally had no explanation (p 62). When originally question about what he did with the balls while in the bathroom, McNally stated that he put the bags on the left and used the urinal on the right (59). However, the bathroom involved did not have a urinal. There were also bathrooms in the officials locker room where the balls were that could have been used. In court, liars often get tripped up on minor details and the I believe the incorrect reference to the urinal, is one of those type of details.
Second. After the deflator texts became available, McNally refused to allow himself to be questioned again about those texts and his role in preparing footballs.
Third. Jastremski, who was also involved in the preparation of the footballs, appears to have lied about receiving a very valuable autographed football from Brady, stating that he received just “just a general football” when he actually received the football that Brady was using when he surpassed 50,000 yards of passing.(p93) A number of text messages later showed how excited Jastremski was to receive the ball.
Fourth, the Patriots decided not to appeal the NFL’s ruling. I believe that if the Patriots truly believed that their personnel were innocent, they would fight the ruling like cats and dogs. The Patriots have no reason to throw Brady under the bus. The fact that they are not appealing indicates to me that there are a lot of skeletons in the closet on this matter.
The bottom line is that whether the balls were deflated or not is a very simple matter. If I had not deflated the balls I would want all of the evidence to come in because I would know that it would tend to exculpate me. In particular, if I was McNally, I would jump at the opportunity to further explain the deflator remarks. Exactly the opposite happened. The fact that so much evasion occurred and the people involved didn’t want to fully open their records is strong evidence to me that the balls were deflated. Also, if McNally had a good reason for taking the balls out of the officials locker room, he should have been able to explain why he did immediately after the game, but he couldn’t do so.
JD
• mpainter'm
Posted Jul 8, 2015 at 11:30 AM | Permalink
JD,
All hinges on whether the Patriots’ balls were deflated. If there was no crime, then there was no culprit.
The points you raise have been rebutted by the Patriots or others, and satisfactorily,
imo.
• jddohio
Posted Jul 8, 2015 at 11:37 AM | Permalink
MP Then go ahead and explain why McNally took the balls out of the officials locker room and had no explanation for it. Why didn’t McNally appear for a second time to explain the deflator texts.
I should add that I personally like Brady. I remember when he was at Michigan and had kind words for Steve Belisari, Ohio State’s quarterback. (I am an OSU fan)
JD
• Steve McIntyre
Posted Jul 8, 2015 at 12:07 PM | Permalink
Why didn’t McNally appear for a second time to explain the deflator texts.
The Patriots (see here) say that McNally was interviewed four times, three times by NFL security and for an entire day by Wells. The Patriots say that Wells’ request for a further interview was outside the scope of their agreement and that Wells did not respond to their request for an explanation.
It sounds to me like the Patriots regarded Wells by this time more as opposing counsel seeking additional discovery beyond the agreed discovery and, presumably like many lawyers in civil litigation, used the occasion to draw a line. While Wells may have been acting like opposing counsel, it was obviously unwise for Patriots not to go the extra mile, especially with someone as antagonistic as Wells was going to be writing the report. I’m not sure that you can draw substantive conclusions from the exchange, other than relations between the lawyers had broken down by this time. Also, the Patriots also presumably believed Brady when he denied any deflation scheme, the existence of which remains dubious to me.
• jddohio
Posted Jul 8, 2015 at 1:10 PM | Permalink
Steve: “It sounds to me like the Patriots regarded Wells by this time more as opposing counsel seeking additional discovery beyond the agreed discovery and, presumably like many lawyers in civil litigation, used the occasion to draw a line. While Wells may have been acting like opposing counsel, it was obviously unwise for Patriots not to go the extra mile, especially with someone as antagonistic as Wells was going to be writing the report.”
In isolation, your statement about this precise issue is reasonable, and the refusal to answer further questions could be viewed as insignificant. However, when placed in the context of McNally not being able to explain why he took the balls out of the officials’ locker room and almost certainly lying about what motivated him to go to the bathroom, I believe the refusal to answer the question is significant evidence against the Patriots. It is part of a bigger pattern of evasion and unexplained breaches of protocol that point the finger at the Patriots. I also agree that generally Wells was not acting in a neutral manner. However, if I didn’t deflate the balls, I wouldn’t care and would want to further explain my position.
JD
• Steve McIntyre
Posted Jul 8, 2015 at 2:43 PM | Permalink
However, when placed in the context of McNally not being able to explain why he took the balls out of the officials’ locker room and almost certainly lying about what motivated him to go to the bathroom, I believe the refusal to answer the question is significant evidence against the Patriots.
McNally’s explanation according to the Wells Report as to why he left with the balls when he did was:
According to McNally, when the NFC Championship Game ended shortly after the start of the overtime period, an unidentified NFL official said something like “we’re back on again,” so he picked up the balls and began to walk out of the Officials Locker Room.
The entire officials’ room emptied within a few minutes of the end of the NFC game, so I don’t see precisely what’s implausible about this aspect of his story.
Nor do we know that McNally was “almost certainly” lying about why he went to the bathroom. I don’t see any evidence that he was lying unless one has independently established that he deflated the footballs. Without being able to establish deflation, I don’t see how you establish lying, so I don’t agree that the alleged lying can be evidence of deflation.
Another point as I re-read the comments about this aspect of this incident. It seems to me that NFL officials embellished their claims to have normally maintained supervision of the balls and that it is likely that McNally’s claim that it was not unusual for him to take the balls to the field unaccompanied is true. Two witnesses supported McNally’s claim on this point. For example:
Paul Galanis, who was stationed just outside the entrance to the Patriots locker room, across the corridor from the top of the center tunnel, said that it was routine for McNally to walk to the field with the game balls unaccompanied. He estimated that McNally goes to the field approximately 10% of the time with game officials and approximately 25-30% of the time with Richard Farley, and the other times he is walking by himself.
NFL security representative Farley’s story was the opposite. The Wells Report says:
he[Farley] considers it part of his job description to accompany the referee to the field and that he is generally in close proximity to McNally and the game balls when he walks to the field with the referee. According to Farley, he often opens the door to allow McNally to exit easily with the ball bags, and then McNally, Farley, the referee and the head linesman will walk to the field together or in close proximity to each other. Farley cannot recall McNally previously bringing game balls to the field prior to the start of a game without being accompanied by or in close proximity to one or more game officials.
I can see why Farley would be engaged in full-scale CYA and his statements sure sound like CYA to me.
Given the lacklustre performance of NFL officials in half-time measurements – when they couldn’t even keep track of which gauge they were using – I don’t see why very much weight should be given to CYA statements as opposed to third party evidence from security guards.
• Steve McIntyre
Posted Jul 8, 2015 at 2:50 PM | Permalink
A further gloss on related points on re-reading the Wells report on this timeline. Unless one watches the pea carefully, one can easily get the impression that McNally was not on the field when Farley arrived on the field shortly after 6:36:
As soon as he reached the field, Farley looked for McNally by the instant replay booth, where McNally regularly arrives with the game balls, but did not see him.
HOwever, elsewhere in the report, they say that McNally had “exited the bathroom at approximately 6:32:27 p.m., and took the bags of game balls to the field.” So McNally was on the field for several minutes by the time that Farley arrived. Precisely why Farley missed him is unexplained in the report, but it was not because McNally wasn’t on the field.
• jddohio
Posted Jul 8, 2015 at 3:02 PM | Permalink
Steve: Here is what McNally stated in his first interview before he had time to think up explanations that could more strongly support his claim that he had done nothing wrong.
(p58)”According to the interview report from a telephone interview with NFL Security on January 19, McNally stated that he stopped to
use the bathroom on the way to the field and took the game balls with him into the bathroom.
59
During this interview, he explained that he did not use the bathroom in the Officials Locker Room because he did not want to disturb the officials. He claimed that he had left the Officials Locker Room with game balls but without a game official on a few occasions over the years, but
could not identify any particular games where that had occurred. According to the interview report from an NFL Security interview of McNally on January 21, McNally said that he did not know why he would leave the locker room with the game balls without being accompanied by game officials, and “just decided to leave the locker room at that time to go to the field.” He said that no one had ever told him that he was required to wait for the officials. He also claimed that he went into the bathroom with the game balls because when he got to the end of the tunnel, he
realized that he suddenly had to use the bathroom.”
I believe his first statement explaining what happened is by far the most important one and that when he could not explain right after the game why he would leave the locker room with the balls while not being accompanied by the officials it is very damaging. Later explanations don’t cut it with me. Also, because he said the bathroom had a urinal when it didn’t, I don’t believe anything he said about his reasons for going to the bathroom. This is not the type of detail you would forget in a couple of hours.
JD
• Posted Jul 8, 2015 at 4:31 PM | Permalink
JD, re: your statements about McNally’s answers in his first interview – the quotes you are citing are from his 2nd and 3rd interviews. His first interview was the night after the game – Jan 18. you are referencing a phone interview Jan 19 (2nd interview – the next day) and another Jan 21 (3rd interview – 3 days after the game). So his recall is not from hours after the game.
I don’t see how you can use his answers against him. In his mind, he didn’t need permission to walk the balls to the field. If his claim (supported by 2 security people who work Pats games) is that he walks the balls to the field unattended 50% of the time, then why would he be doing anything different from what he normally does – there’s no need to answer why he decided to walk the balls down to the field unaccompanied, it’s what he does at least 50% of the time.
The AFC game start was delayed because the NFC game went to OT, so everyone was on hold until it ended. McNally was waiting for the NFC game to end so he could walk the balls to the field after getting permission from Anderson to remove the balls from the officials room. The NFC game ends, which means the AFC game prep is back on again, and he walks the balls down to the field – walking right through the crowded officials room with 2 bags of 12 balls each over each shoulder. Walking right past the vision of a high level NFL who knew about the deflation accusations prior to the game. He walks by everyone, down the tunnel, stops to use the bathroom on his way because he has to go, and continues on to the field.
If walking the balls unaccompanied was his normal routine more often than not, and he’s been doing this for years (he’s been a ball boy or locker room attendant for 32 years), then I would argue he’s on auto-pilot for that part of his job. He also said he routinely uses the bathroom on the way to the field. When I walk in the office every am, I go from my car to the front office door and then I walk back to my office. If you asked me the details of this 2 days later – did I stop and talk to someone on the way back to my office, did i use the bathroom on the way to my office, or after I put my stuff in my office or not, etc. I wouldn’t be able to tell you what i did or didn’t do. (side note – Hopefully there was no crime committed at my office 2 days ago, because I wouldn’t come across as being forthcoming, lol).
• mpainter'm
Posted Jul 8, 2015 at 5:31 PM | Permalink
McNally’s account gives a picture of fuzzy-wuzzy protocol and lax procedures, to the discredit, ultimately, of Roger Goodell.
Ted Wells has the task of painting McNally as a liar as well as a culprit, unless he wants a dissatisfied client. Unfortunately for Wells & Co. and Goodell to boot, there is eyewitness corroboration of McNally’s account and also Mr. Farley has been sniffed out in his clumsy attempt to color the situation. When the book is written, Wells, Exponent, Goodell and their minions are all going to appear as fumbling clowns.
Why was John Raucci, Chief Investigator for the NFL, at this game? What was he investigating?
• chuckrr
Posted Jul 8, 2015 at 7:45 PM | Permalink
“Unless one watches the pea carefully, one can easily get the impression that McNally was not on the field when Farley arrived on the field shortly after 6:36:”
This point by Steve illustrates the biased nature of the Wells Report. McNall was on the field yet Farley couldn’t find him. it’s really a completely irrelevant point but somehow it makes it into the narrative of the story. But notice how you (or Steve in this case) have to put two and two together That Farley just didn’t see him. The implication…unsaid…McNally is hiding from the officials. It’s this way throughout the report.
• Carrick
Posted Jul 8, 2015 at 9:24 PM | Permalink
Steve McIntyre:
The Patriots (see here) say that McNally was interviewed four times, three times by NFL security and for an entire day by Wells. The Patriots say that Wells’ request for a further interview was outside the scope of their agreement and that Wells did not respond to their request for an explanation.
The Patriots know bloody well that McNally was interviewed exactly one time by the NFL investigative team. As JD points out, it is more likely in their economic interests to not be fully forthcoming, whereas it’s hard to understand why the game officials would be less than as honest.
I see an undue amount of weight being put on “what the Patriots say”, especially in cases like this when what they are saying is such obvious spin.
HOwever, elsewhere in the report, they say that McNally had “exited the bathroom at approximately 6:32:27 p.m., and took the bags of game balls to the field.” So McNally was on the field for several minutes by the time that Farley arrived. Precisely why Farley missed him is unexplained in the report, but it was not because McNally wasn’t on the field.
This was addressed in the report.
As soon as he reached the field, Farley looked for McNally by the instant replay booth, where McNally regularly arrives with the game balls, but did not see him.
The fact that Farley didn’t find McNally is just further evidence there was something very usual in McNally’s behavior. This combined with his inconsistent statements seems like a major red flag to me.
• MikeN
Posted Jul 9, 2015 at 7:21 AM | Permalink
Carrick, that is the discrepancy. The Patriots are claiming the video shows McNally going to the field from the tunnel and being in his usual place. Wells Report has Farley leaving the tunnel and looking in that same place, presumably also from video. Why didn’t Farley see him?
• Steve McIntyre
Posted Jul 10, 2015 at 1:10 PM | Permalink
Carrick writes:
As soon as he reached the field, Farley looked for McNally by the instant replay booth, where McNally regularly arrives with the game balls, but did not see him.
The fact that Farley didn’t find McNally is just further evidence there was something very usual in McNally’s behavior. This combined with his inconsistent statements seems like a major red flag to me.
Carrick, there seems to be evidence that McNally went to the replay booth as he was supposed to, evidence that was not mentioned by the very biased Wells Report, leading you to view something as deceptive that wasn’t.
The PAtriots say:
Once the footballs are taken to the field they are to be taken to the area adjacent to the replay booth. The outdoor security camera shows that is exactly what Mr. McNally did. Anyone actually concerned about the location of the game footballs could simply have checked that location.
The security video shows Mr. Anderson coming out to the field and going there. Not surprisingly, he found Mr. McNally was there with the bags of footballs. No one then reprimanded Mr. McNally for having taken the footballs without permission or accompaniment, although the report would have one now believe that officials thought Mr. McNally had done something wrong by taking the footballs himself.
I know that you disbelieve Patriot statements, but here they adduce security camera footage as evidence. While I haven’t seen the security footage, it doesn’t seem to me to be the sort of claim that they would make if it weren’t supported by the security footage. The security footage was available to the Wells Report and they did not say that it shows something else.
Now here’s how the Wells report describes events:
As soon as he reached the field, Farley looked for McNally by the instant replay booth, where McNally regularly arrives with the game balls, but did not see him. He did, however, see John Raucci, Director of Investigative Services at the NFL, shortly after stepping onto the field and asked if Raucci had seen either McNally or the game balls. Raucci said that he had seen neither. Farley headed back toward the Officials Locker Room to get the back-up balls. He is seen on the security footage at approximately 6:42 p. m. walking back down the tunnel leading to the field with the bags of back-up balls.
Shortly after taking the field, after Farley had returned to the Officials Locker Room for the back-up balls, Anderson and the other officials noticed that McNally and the game balls were on the field.
In this rendition, there is no mention of where McNally and the balls were located, which, according to the Patriots’ account of security footage, was right at the replay booth where he was supposed to be all along, and apparently was.
That the Wells Report’s account convinced someone as able as you that this seemingly nothing aspect of this incident was “a major red flag” and showed something “very unusual” only proves to me that the Wells Report wrote up this aspect of the affair very deceptively.
• Carrick
Posted Jul 9, 2015 at 9:58 AM | Permalink
MikeN, he eventually took his usual place. I think even McNally’s testimony has him going over to the Patriots bench first. Of course New England knows this but chooses to obfuscate the issue in their rebuttal. Why do you suppose they did that?
• Carrick
Posted Jul 10, 2015 at 2:32 PM | Permalink
I’ve looked at this and found it much less satisfying than you apparently do.
It’s unfortunate the Patriots don’t release the security film (which hopefully has time stamps on the frames). I am skeptical of the motives behind the releaing of a verbal description when videographic evidence exists.
Of particular issue here is that Patriots don’t specifically say that McNally was at the replay booth when Farley arrived looking for him, but they do confirm that McNally was there once Anderson arrived. Curious omission.
Nor do they provide a timeline stating the times when McNally arrived as opposed to when Farley arrived at the replay booth. Given that the Wells report does include times, when available, this is another curious omission.
The question was always “was McNally at the replay booth when Farley arrived”, a question that is should have been directly addressed, but likely deliberately avoided, in the Patriots response:
Once the footballs are taken to the field they are to be taken to the area adjacent to the replay booth. The outdoor security camera shows that is exactly what Mr. McNally did.
Yes, McNally did (eventually) arrive at the replay booth. That is not even a point in dispute.
The question that actually is in dispute is “when did McFarley arrive?” Why exactly is this critical information omitted and how can you be satisfied with a response that actually fails to provide key information such as this?
The Wells report states very clearly that Farley went to the replay booth. They state he did not see McNally there.
They state that John Raucci, who was also at the replay booth, also did not know where McNally was. If McNally was at the booth at that point, Raucci must have seen him arrive.
I find it implausible that McNally was at the booth when Farley arrived there. I would be willing to accept documentary evidence to the contrary. Although New England had access to documentary evidence that could have contradicted this conclusion, they withheld this information, and only provided ancillary and noncontroversial information instead.
And you find that satisfying?
When I read the Wells report, I see evidence that is exculpatory being reports (such as the security guards account on McNally). When I read the Wells report, a timeline is provided. When I the Patriots report, details that matter are left out, while points that are not in dispute get amplified. The document that is dropping critical quantitative details while liberally using spin appears to me to be the Patriots response.
Yet it’s the Wells report that I am supposed to find deceptive here—even though neither the Wells commission nor the NFL has anything to gain by smearing the Patriots organization or one of its heroes, Tom Brady. On the other hand, the Patriots response, which almost definitionally is a self-serving document, apparently I’m to view uncritically and as a paragon of the truth.
• MikeN
Posted Jul 10, 2015 at 6:45 PM | Permalink
Carrick one explanation is that McNally went somewhere else first and then arrived at the replay booth, not described anywhere I’ve seen, but does fit the facts.
Some other possibilities:
Maybe there is a replay booth and an instant replay. Wells said instant replay.
Farley never actually went to the replay area. Wells said he looked there, but that could mean he came to the field and turned in that direction. It also doesn’t say he saw Raucci at the replay booth either, but shortly after he(Farley) stepped onto the field. It’s not clear where the tunnel exits onto the field.
• chuckrr
Posted Jul 10, 2015 at 7:43 PM | Permalink
Did I miss something? Is there a point to this latest round of inquiry? There’s only one place he had the time to deflate the balls. That was in the 9×9 bathroom with a toilet and a sink and two bags of balls by a huge man in under 100 seconds.
• mpainter'm
Posted Jul 8, 2015 at 12:05 PM | Permalink
JD
Read the comments. All of these issues are explored and clarified. Also, there are one or two links to the Patriots’ rebuttals on issues raised in the Wells Report.
Please excuse me from a discussion of this tangle.
• JD Ohio
Posted Jul 8, 2015 at 12:10 PM | Permalink
MP. If you are going to comment, you should be able to respond. Read the comments is not an answer or response.
JD
• mpainter'm
Posted Jul 8, 2015 at 2:25 PM | Permalink
I can comment on the science;
that I understand. I simply referred to better sources than I on your inquiries.
I have a question for you:
You have a legal B/G, if I’m not mistaken.
Does this work by Ted Wells and his associates enjoy privilege? I ask because it deals not with matters of the law but only with NFL regulations.
• jddohio
Posted Jul 8, 2015 at 9:25 PM | Permalink
M Painter “Does this work by Ted Wells and his associates enjoy privilege?”
You are asking a very subtle question that I can’t answer. The attorney client privilege is structured to deal with the typical adversarial situation where opposing attorneys clearly represent their own clients. In this instance, Wells is purportedly acting as an impartial investigator. How that would play out in terms of attorney client privilege I don’t know. More importantly, I would expect that the NFL charter or articles of incorporation spell out what portions of an investigator’s work are proprietary to the NFL and which parts would be open to all parties. (This is also speculation on my part)
JD
• MikeN
Posted Jul 10, 2015 at 10:11 AM | Permalink
Carrick, I don’t see McNally testimony as to going to the Patriots bench first, but that would explain the discrepancy.
• MikeN
Posted Jul 8, 2015 at 1:40 PM | Permalink
JD regarding the football that Jastremski lied about, that is a separate issue entirely. It appears he lied to various people and gave it more importance than it actually had, and didn’t want to admit to the lie he told to his family and friends. I think the investigators would have been better off leaving that out of the report.
• chuckrr
Posted Jul 8, 2015 at 5:13 PM | Permalink
I think I saw some texts also that may have caused some trouble for Jastremski with his wife.
14. Posted Jul 8, 2015 at 3:33 PM | Permalink
JD, I disagree with your statement “McNally not being able to explain why he took the balls out of the officials’ locker room and almost certainly lying about what motivated him to go to the bathroom, I believe the refusal to answer the question is significant evidence against the Patriots.” I also disagree that he was not being cooperative.
Here’s how the interview process went with McNally according to the Patriots, he seems cooperative IMO. from http://wellsreportcontext.com/#patriotscooperation :
The first of Mr. McNally’s interviews happened the night after the game, when Mr. McNally volunteered to stay at the stadium for an interview since he would not be back for his game-day responsibilities until August. Patriots management had not yet been advised that an investigation had started, but Mr. McNally, having nothing to hide, talked freely to the League personnel without even asking if someone from the team should be there with him. The second and third interviews happened within the next several days. Again, Mr. McNally gave these interviews without any Patriots representative with him. His phone was offered to League personnel for imaging, but they advised that they did not need his phone. (His phone data was later provided to the Wells investigators upon their request and prior to their interview with him.) At his third interview with League Security personnel, he was subjected to very aggressive questioning and demeaning assertions that he was lying when he denied any knowledge of improper football deflation. This approach to the issues by League personnel was consistent with their prejudgments of wrongdoing by the Patriots. Notwithstanding that he had already been interviewed three times, when Mr. Wells asked to interview him again, the Patriots agreed to facilitate that fourth interview. That agreement was based on an explicit understanding reached with the Wells investigators: barring unanticipated circumstances, individuals would only be interviewed by the Wells investigators one time.
Here’s some context for McNally’s initial answers, where you say he didn’t have an answer for why he took the balls:
League security actually began investigating during the second half of the game when they began questioning Patriots ball boys. Consistent with that, Mr. McNally described the focus of his first interview as being on the role of ball boys. It was accurate for him to have stated that nothing unusual happened during the walk from the locker room to the field, since, as he later explained, his bathroom stop was nothing unusual. When later asked why he did not use the urinals in the Officials’ Locker Room or the chain gang room, he fully explained why — and his reasons are supported by the report’s conclusions about how crowded the Officials’ Locker Room area was (pgs. 54-55). One can draw no adverse inferences from an attendant deciding not to use the crowded facilities. If the investigators had found a single witness who had seen Mr. McNally routinely using the urinals in the Officials’ Locker Room prior to other games when the officials were doing their final pre-game preparations, they would have put that in the report. The bathroom he used is on his direct route from the Officials’ Locker Room to the field.
Here’s more context for McNally taking balls to the Field:
What Mr. McNally actually described was exactly what the report stated happened before the AFC Championship Game — that he gets permission from the game officials to remove the footballs from where they reside in the dressing room of the Officials’ Locker Room. As the report acknowledges (pg. 55), Mr. McNally received precisely that permission: “Anderson also recalls that Mr. McNally, with Anderson’s permission, had moved the bags of footballs from the dressing room area towards the sitting room shortly after the officials returned from the player’s walk-through.” Thus, Mr. McNally had the referee’s permission to remove the footballs from the part of the dressing room where game officials congregate pre-game. He sat with the footballs in the sitting room and then, when the NFC Championship Game that everyone was watching in that sitting room ended, he took the footballs from the sitting room and out into the hallway in full view of numerous League and game officials. Even after halftime, when psi measurements had become an issue, Mr. McNally is seen on the security tape walking the footballs back to the field totally unaccompanied by any game or League official, but obviously with their full knowledge that he was doing so. Again, no one told him to wait, to stop, or that he was doing anything wrong in taking the footballs from the Officials’ Locker Room to the field.
From the Wells Report: With respect to his decision to leave the locker room unaccompanied, McNally claimed that his actions on the day of the AFC Championship Game were not unusual. In his account, the game balls remainin the locker room until he believes it is time to take them to the field. According to McNally, he brings the game balls to the fieldwhen he deems fit. He said that he generally asks permission or alerts the officials before he moves the game balls from the dressing room to the sitting room, but does not ask or alert them again before leaving the locker room and taking the balls to the field. McNally also claimed that it is not his customary or typical practice to walk to the field with game officials. He estimated that he walks to the field with other people—game officials, security personnel or others—only half of the time. The Patriots produced two game-day security guards employed by Team Ops, a
security and guest services company affiliated with the Patriots, to support McNally‟s account. Rita Callendar, who was stationed just outside the Officials Locker Room on game day, said that she estimates that McNally takes the game balls to the field by himself roughly 50% of the time, and that the other times he walks with or in close proximity to Richard Farley. Paul Galanis, who was stationed just outside the entrance to the Patriots locker room, across the corridor from the top of the center tunnel, said that it was routine for McNally to walk to the field with the game balls unaccompanied. He estimated that McNally goes to the field approximately 10% of the time with game officials and approximately 25-30% of the time with Richard Farley, and the
other times he is walking by himself.
Here’s the Patriots version of him walking the balls out of the locker room, all seen on the security video:
When the NFC Championship Game ended abruptly in overtime and Mr. McNally started from the back of the sitting room towards the door to the hallway, he walked by numerous League officials in the sitting room. As the report states (pg. 55), the sitting room was crowded with “NFL personnel, game officials and others gathered there to watch the conclusion of the NFC Championship Game on television.” Mr. McNally had to navigate this crowd of officials to make it through the sitting room with two large bags of footballs on his shoulders. Mr. McNally walked past all these League officials and out the door of the Officials’ Locker Room. As is clear from the report, no one objected; no one told him to stop; no one requested that he wait to be accompanied by a League official; no one told him that a League official had to carry the footballs to the field. After he walked past all of these League officials and out the door of the Officials’ Locker Room to the hallway, he then walked past James Daniel, an NFL official and one of the people who had been alerted to the Colts psi concerns pre-game (pg. 45). Mr. Daniel, as seen on the security video, looked at Mr. McNally carrying the bags of footballs toward the field unaccompanied by any League or game official, and made no objection to Mr. McNally continuing unaccompanied to the field. If officials lost track of the location of game footballs, it was not because Mr. McNally stealthily removed them. (Omitted from the investigation were interviews with all those League officials whom Mr. McNally walked past with the bags of footballs on his shoulders.) Even after halftime, when obvious attention was being paid to game footballs and psi issues by League and game officials, who took control of the footballs at halftime, the security video shows Mr. McNally, with no objection, taking the footballs from the Officials’ Locker Room back to the field totally unaccompanied by any League or Game official.
• jddohio
Posted Jul 8, 2015 at 3:57 PM | Permalink
D 1964
Much of the Wells report contradicts the essence (not the literal truth) of your description. For instance: “Anderson said that it is typical for locker room attendants throughout the League to help move the game balls towards the front of the locker room, but that the footballs do not leave the locker room until the officials give express permission for them to be brought to the field at or near the time the officials also walk to the field. Numerous other game officials described a similar practice.”
****
“Both Anderson and Veteri immediately asked Farley where the footballs were. Farley checked for the ball bags in the back part of the locker room (where he saw the bags of back-up balls) and in the adjacent Chain Gang Locker Room, but could not find them. When it was suggested that McNally had or may have taken them to the field, Anderson responded that “he‟s not supposed to do that.” Anderson also stated that “we have to find the footballs.” Blakeman recalls that although Anderson is usually
calm and composed leading up to a game, Anderson was visibly concerned and uncharacteristically used an expletive when the game balls could not be located. The other officials were similarly surprised and concerned. None of the officials in the locker room at the time realized that the game balls had been removed from the locker room until they were ready
to go to the field for the start of the game, and all expected that the balls would not leave the locker room until it was time for them to take the field.”
Part of what persuades me that the officials version was correct was McNally’s inability at the first interview to identify any game where he had taken the game balls without accompanying the officials. Also, the fact that he apparently came up with an explanation of what he did with the balls after the first interview is not very impressive to me The fact that he walked by numerous NFL officials with the balls doesn’t impress me much if they were not charged with supervision of the balls. (If he had walked directly by one of the officials described above that would be a different matter.)
My basic point, which does not appear to be significantly contradicted by you, is that where there is a conflict in explanations and statements, I give the most credence to the initial statement. I would also add that what is happening between us is precisely what happens at jury trials. You are essentially giving no credence to the officials and I am essentially giving no credence to the Patriots employees later explanations of what occurred. If actual transcripts of the interviews were provided, some of this would probably be cleared up.
JD
• Posted Jul 8, 2015 at 5:36 PM | Permalink
JD, I agree that we really need the transcripts (video of interviews would be even better).
I don’t think the conflicts in statements necessarily point to guilt.
I see it this way – you have a low level employee who immediately after the game is questioned with the questions focused on the ball boys and whether anything unusual had taken place. He volunteers to answer questions that night instead of coming back the next day. He does so without any patriots personnel present. Those actions are consistent with an innocent person. When he is answering questions that night, he has done nothing earlier that day he hasn’t normally done for many games over the years (his assertion that he does this 50% of games jives with 2 other security people who work pats games) Just because Anderson and some of the other league refs don’t allow walking to the field unaccompanied, doesn’t mean others don’t). So when they ask about anything unusual, walking the balls to the field unaccompanied and using the bathroom is not unusual behavior to Mcnally – so he doesn’t say anything about it.
So first interview – the question was, “did you do anything unusual?” McNally doesn’t think using the bathroom on the way to the field is unsual, so he doesn’t mention it.
2nd or 3rd interview – NFL sees he used the bathroom, so they ask him about it. Again, if it’s something he’s done in the past and isn’t unusual (he’s worked for the pats for 32 years, not sure how old the stadium is, but it’s a familiar bathroom to him), the details wouldn’t stand out to him. I can see someone calling a toilet a urinal or vice versa
I would argue if he was using the bathroom to deflate balls, he’d be on high alert to get it done without getting caught. I think he’d remember every last detail of the bathroom. The fact that he has a vague recollection of it and can’t remember where he put the balls, etc. points to it being such a non-event that he’s on auto-pilot, where he’s going through the motions but his brain isn’t registering the details.
He also volunteered his phone during one of the interviews – the NFL declined.
But your right – seeing the transcripts (video would be best so we could see tone and body language) would give us context to the exact questions and answers, where we could make a better determination if he was being truthful or evasive.
• MikeN
Posted Jul 8, 2015 at 7:48 PM | Permalink
JD, assume that McNally is right and he has taken footballs to the field unaccompanied before, maybe a quarter to half the time as the security guys said. Now when asked, would you expect him to identify a specific game where he did so?
• jddohio
Posted Jul 8, 2015 at 9:21 PM | Permalink
MikeN “Now when asked, would you expect him to identify a specific game where he did so?”
Yes, that should be quite easy. Suppose he remembers 3 games and is wrong about one of them. No big deal. He still can show that he took the balls out on his own accord in 2 games.
JD
• MikeN
Posted Jul 9, 2015 at 2:23 PM | Permalink
I wouldn’t expect to remember specifically in this game I took the balls by myself, while in this game the refs came with me.
• MikeN
Posted Jul 8, 2015 at 7:50 PM | Permalink
If the officials never allow this to happen, why did they allow it a second time at halftime?
This was a game where the officials were on notice to be alert with the footballs, they reinflated after busting the Pats’ deflating, then they let McNally take the footballs by himself?
• mpainter'm
Posted Jul 9, 2015 at 8:21 AM | Permalink
JD,
From the Wells Report:
“Rule 2….and the balls shall remain under the supervision of the referee until they are delivered to the ball attendant just prior to start of the game.”
Nothing about the referee escorting the ball attendant onto the playing field. I imagine that other NFL ball attendants would give accounts very similar to McNally’s.
Now, I’m sure that Wells and his crew can read just as well as you or I.
So why does he pretend that it is an issue? Because it is not.
The Wells Report should be taken for what it is.
• MikeN
Posted Jul 9, 2015 at 4:32 PM | Permalink
>If actual transcripts of the interviews were provided, some of this would probably be cleared up.
I think the tunnel video is sufficient. After looking at a picture of McNally, I feel confident that he would not be able to deflate 13 footballs in that timeframe, or if he somehow did, he would not be relaxed but instead breathing heavily.
15. MikeN
Posted Jul 8, 2015 at 3:49 PM | Permalink
JD I see nothing wrong with McNally’s postgame answers that lead me to think of him as guilty. He couldn’t remember a prior game that he took the footballs is not the same thing as he never left with the footballs before. He just can’t identify a specific game. He has no explanation for why he left without game officials because no one told him he couldn’t do that before.
For that matter the referees had him at halftime take the balls out unaccompanied, right after they caught the Patriots cheating and reinflated them.
• jddohio
Posted Jul 8, 2015 at 4:05 PM | Permalink
MN You are totally disregarding the officials statements that the balls are to be taken to the field with the officials present. That is your right. But between the officials (each of whom have other full time jobs) and McNally, I would tend to believe the officials particularly when McNally originally had no explanation as to why he would transport the balls without the officials present.
JD
• Posted Jul 8, 2015 at 4:35 PM | Permalink
JDOhio, what about the 2 security officials the pats made available – they both said mcnally took balls unaccompanied 50% of the time.
Also, the officials work for the NFL – could there be a potential that refs are giving an “answer” based on what they are supposed to do, vs. giving an answer that may expose that they are not following the league protocols?
• jddohio
Posted Jul 8, 2015 at 4:46 PM | Permalink
D 1964
One way to resolve much of this issue is to ask other officials and staff members for other teams what the policy is. I don’t think everyone would lie. Why (to my knowledge) hasn’t anyone outside of the Patriots stated that it was not unusual for team employees to take the balls to the field unaccompanied. As you can tell, I don’t give much credence to the Patriots statements. If I was an official, I wouldn’t lie about procedures and depend on the other [roughly 70] officials to back up my lie. On other hand, if I was a Patriots employee, I could infer that it is in my economic interest to lie.
JD
• MikeN
Posted Jul 9, 2015 at 8:45 AM | Permalink
It’s not a lie. They probably think the procedure is to accompany the footballs. However, during the various games they probably don’t care about it and just have the guy go out there by himself, to the point where maybe they are only there half the time as the security guys said.
• chuckrr
Posted Jul 8, 2015 at 5:36 PM | Permalink
JD
From the Wells Report…according to the officials
“McNally had not previously removed game balls from the Officials Locker Room and taken them to the field without either receiving permission from the game officials or being accompanied by one or more officials.”
Notice the part about receiving permission. McNally thought he had permission. maybe he was wrong. but it appears that there wasn’t really much focus on this part of the procedures up until this game.
16. MikeN
Posted Jul 8, 2015 at 3:51 PM | Permalink
> After the deflator texts became available, McNally refused to allow himself to be questioned again about those texts and his role in preparing footballs.
MaNally was questioned after those texts became available. The investigators just didn’t ask about it because they hadn’t noticed them.
• chuckrr
Posted Jul 8, 2015 at 5:19 PM | Permalink
It needs to be noted….again… that there is one deflator text. Only one and it was on May 9th in the off season. And they were not talking about footballs
• MikeN
Posted Jul 8, 2015 at 7:41 PM | Permalink
There is another one that says deflate and give somebody that jacket. Not highlighted as much since it supports the theory of weight loss. However it was at halftime against Green Bay when they were down nine.
• chuckrr
Posted Jul 8, 2015 at 8:01 PM | Permalink
Is that in the Wells Report?
• MikeN
Posted Jul 9, 2015 at 7:37 AM | Permalink
It was a road game, so again McNally had no chance to deflate, plus Wells Report as it as being sent by McNally to Jastremski. No explanation was given by Wells Report as to why McNally wanted Jastremski to do McNally’s job.
• MikeN
Posted Jul 9, 2015 at 7:30 AM | Permalink
There is also “The only thing deflating sun. is his passing rating”
Next one will be a balloon
Make sure you blow up the ball like a rugby ball so tom can get used to it before sunday
Make sure the pump is attached to the needle, watermelons coming
all right after the jets game.
• chuckrr
Posted Jul 9, 2015 at 8:44 AM | Permalink
I’m referring to NcNally or Jastremski calling each other deflator. I realize they used the word in some verb form a few times.
• Carrick
Posted Jul 9, 2015 at 10:09 AM | Permalink
MikeN:
MaNally was questioned after those texts became available. The investigators just didn’t ask about it because they hadn’t noticed them.
Or because they simply assumed they’d have a follow-up interview and were giving him rope to hang himself with. Remember there was a single interview between McNally and the NFL investigative team.
As I mentioned above, I believe this lack of cooperation with the investigative team is what lead to such heavy fines rather than New England potentially modifying the football pressure.
Anyway, if you look at the words surrounding “deflator” it does not seem consistent with discussing weight loss.
Nice dude….jimmy needs some kicks….lets make a deal…..come on help the deflator
(The reply from Jastremski was apparently deleted.)
This is followed roughly eight minutes later by:
Chill buddy im just fuckin with you ….im not going to espn……..yet
So by you guys reading, McNally is threatening to go to espn to talk about his weight loss?
• Steve McIntyre
Posted Jul 9, 2015 at 2:07 PM | Permalink
Carrick, you say:
Or because they simply assumed they’d have a follow-up interview and were giving him rope to hang himself with. Remember there was a single interview between McNally and the NFL investigative team.
The Patriots put their correspondence with the league about interviews online here. On February 5, the Patriots and the NFL agreed as follows:
Scheduling of witness interviews: You have requested new
interviews of those already interviewed, as well as interviews of a
number of other individuals. We will work to accommodate all
those interviews. The interviews will be arranged so
that, barring unanticipated circumstances, there will
not be future multiple interviews of the same person
A subsequent letter in March says that by that time McNally had been interviewed three times by NFL security and once by Wells and sought an explanation of the “unanticipated circumstances” that required a fifth interview. Their letter also complained that League security had leaked McNally’s name to the media and that McNally, a minimum wage employee, and his family were being harassed.
While I think that the Patriots were unwise not to go the further mile, I can also see why they were annoyed and why they felt entitled under their agreement with the NFL to ask for an explanation of the “unanticipated circumstances”.
• Carrick
Posted Jul 9, 2015 at 4:16 PM | Permalink
Steve McIntyre:
While I think that the Patriots were unwise not to go the further mile, I can also see why they were annoyed and why they felt entitled under their agreement with the NFL to ask for an explanation of the “unanticipated circumstances”.
Obviously I’m not very sympathetic on this point:
There was exactly one interview with the NFL investigative group. That is indisputable. The team could and would have traveled to meet McNally so the claim of hardship must be met with due skepticism.
Denying them a second interview wasn’t just foolish, it was actually a breech of contract issue.
As I mentioned above, it’s my understanding that most of the penalties were incurred due to a failure to comply with the terms of their contract rather than the supposed tampering with the Patriots footballs.
The deflator and espn references are discussed by the Patriots as follows:
What was really meant is obviously left up to McNally, not to the Patriots self-servering spin control central. So the answers must come directly from McNally and not some intermediary.
However, the NFL investigative group was not given an opportunity to follow up and ask any further questions on this issue, so we’ll never know.
I certainly don’t accept the Patriots self-serving account on this issue. Why should I believe them and why do you?
• MikeN
Posted Jul 9, 2015 at 7:27 PM | Permalink
It’s not a matter of believing them but seeing if it fits the evidence. My immediate reaction to the security guards’ claims about McNally was that it was people the Pats put forward so no reason to believe them. In this case, deflator as weight loss seemed like a stretch, but then looking at his picture, I can see that it might be used since he is not just big and heavy but close to balloonish. For the Wells team explanation to work, they have to explain why McNally would tell Jastremski ‘deflate and give somebody that jacket.’ Even if they got it backwards, it was a road game, so McNally has no chance to deflate footballs, even though they tried to imply that he did so at halftime.
For the ESPN, again I find it more plausible than the Wells explanation if only because I don’t think ESPN is the name I would use there. My initial reaction when reading the Wells report to this text is that they were deflating Brady’s tires.
Steve: “deflate and give somebody that jkt. [jacket]” was characterized by Wells as follows: “A text message sent by McNally to Jastremski during a Patriots road game against the Green Bay Packers on November 30, 2014, includes a possible suggestion to “deflate” footballs.” They said that they were unable to discuss the meaning of the text with McNally because Patriots’ counsel “refused” to make him available. As the only other usage of “deflate”, this is an important text. AS you point out, it makes no sense to interpret this message as an instruction to deflate footballs when McNally was in a different city.
• Steve McIntyre
Posted Jul 9, 2015 at 11:04 PM | Permalink
A sports blog provided a completely convincing explanation of “deflate and give somebody that jkt” that both showed that jkt meant jacket and that, contrary to Wells, was NOT a “possible suggestion to “deflate” footballs. See here.
The text occurred during a televised game between Patriots and Green Bay. An alert blog commenter synchronized the time of McNally’s text to the televised game and determined that the text immediately followed a Packer touchdown and that, at that precise time, Jastremski was pictured on television holding a jacket, while McNally was at home in New England.
One blogger write:
Some are suggesting it means McNally is referring to the puffy jacket that Jastremski is wearing. Personally, I’m guessing McNally is telling Jastremski to calm down after seeing that play. While that does not refer to weight loss, it does show they loosely use that term for meanings that have nothing to do with air pressure and footballs.
Whatever the precise intent, Wells’ insinuations that this text was a “possible suggestion” to deflate footballs is clearly contradicted.
• MikeN
Posted Jul 9, 2015 at 8:36 PM | Permalink
I don’t see anywhere Wells’s agreeing to the Patriots’ interview terms.
Jastremski was interviewed a second time, and in his second interview nothing new came up. This was a reason for the counsel to not provide McNally. It’s not clear if they asked Jastremski to explain deflator either.
I’m a little confused as to why Wells didn’t go to Roger Goodell to force the issue.
• Steve McIntyre
Posted Jul 9, 2015 at 2:11 PM | Permalink
The deflator and espn references are discussed by the Patriots as follows:
• MikeN
Posted Jul 9, 2015 at 2:20 PM | Permalink
They held off on asking to ask at a later interview? I assume they were given more than five minutes to ask questions.
They missed it.
The explanation isn’t that they were discussing weight loss, but just another nickname.
The ESPN part isn’t with regards to weight loss, but that he was getting free stuff illicitly. Could also be a reference to something still unknown.
If he was engaged in a scheme to deflate footballs, threatening to go to ESPN is not what comes to mind, but to Goodell or even the FBI.
• Carrick
Posted Jul 9, 2015 at 4:22 PM | Permalink
MikeN:
They held off on asking to ask at a later interview? I assume they were given more than five minutes to ask questions.
Why is deferring some questions for a later interview so difficult to accept as a possibility? I don’t see it as implausible.
The ESPN part isn’t with regards to weight loss, but that he was getting free stuff illicitly.
Based on the emails and texts, that’s about half the Patriots staff. Definitely worth a full fledge yawn.
You may decide to credulously accept the Patriots account on this, but I feel no obligation to trust them, especially when they have clearly been obstructionistic towards this investigation.
• mpainter'm
Posted Jul 9, 2015 at 5:24 PM | Permalink
Email from Goldberg to Ted Wells dated 3/9/15:
“If you want some added information from Jim McNally,let me know what it is and I will consider the best way to get relevant information to you”
This is what Ted Wells publicly characterized as a “lack of cooperation from the Patriots”. Wells never availed himself of Goldberg’s offer.
• MikeN
Posted Jul 9, 2015 at 8:38 PM | Permalink
I don’t hold that against Wells, since he considers McNally guilty, and letting him know ahead of time what he wants to ask won’t work. I’m surprised they didn’t request the two guys simultaneously for interviews.
• MikeN
Posted Jul 10, 2015 at 10:45 AM | Permalink
Interview for 7 hours, but decide not to ask about this? It doesn’t take too much to start out with what nicknames do you use for each other, you ever make jokes about deflating footballs, etc.
Highly implausible that they would hold off for a later interview.
Beyond that, Wells has said he didn’t notice the text messages. He was caught in another lie because he said he had only analyzed the previous season to that point. Then in the Wells Report, they said they wanted to ask about deflate and give somebody that jkt, which was from November of the season they supposedly had looked at.
• Carrick
Posted Jul 10, 2015 at 3:12 PM | Permalink
mpainter:
This is what Ted Wells publicly characterized as a “lack of cooperation from the Patriots”. Wells never availed himself of Goldberg’s offer.
Anything Goldberg provided Wells at that point would anecdotal, and there’d be no control over which pieces of information were reported accurately and which contained modified or redacted information.
If Wells had accepted this offer, it could credibly be viewed as a sign-off on the part of the league that by this unacceptable method of communication with McNally, that the Patriots had met their contractual obligations to fully cooperate with the league in this investigation.
Obviously Wells should not sign-off that the contractual obligation had been met, when they hadn’t.
MikeN: Seven hours isn’t all that many for an interview with a key individual in an investigation. You can speculate on why any one point wasn’t discussed, but it’s irrelevant. It wasn’t discussed. Asking for a follow-up interview and expecting that points that weren’t raised (for ANY reason) in the initial interview could be asked in the follow up is completely reasonable here.
The Patriots were required to cooperate in this investigation, but prevented a second interview between the NFL investigative team and McNally. If they were truly innocent of any wrong-doing, they really screwed the pooch on that one.
You are making a big stink of the fact that the point didn’t get asked in the first interview. I see that as a smoke screen to itself cover up obstructionism on the part of the Patriots organization. I think that obstructionism is the real story.
While I would agree with calling ‘enough” if an investigation lingers on forever, this one didn’t linger on, it was cut short.
• MikeN
Posted Jul 10, 2015 at 5:31 PM | Permalink
I think it is reasonable to ask for the followup interview. I don’t think it is reasonable to not bring up something in a seven hour interview, which presumably was open ended.
>You are making a big stink of the fact that the point didn’t get asked in the first interview.
I’m not. The Patriots are saying it.
Obstructionism is correct, and it is why they were punished.
• Carrick
Posted Jul 10, 2015 at 11:31 PM | Permalink
MikeN:
Obstructionism is correct, and it is why they were punished.
I think so to.
I think it would have been a wrist-slap for tampering with the footballs had they cooperated (assuming that was the outcome). In that case, I’ve have expected the major long-term consequence to have been a rule change to permit tweaking the pressure (within allowed ranges) by team representatives.
I think that should have been allowed all along.
17. JonP
Posted Jul 8, 2015 at 5:37 PM | Permalink
“You are totally disregarding the officials statements that the balls are to be taken to the field with the officials present.”
JD you are totally disregarding the video evidence showing McNally did exactly that, twice, in the same game. Who are we to believe video evidence or the refs CYA statements.
Also, Why did the Wells report dismiss Anderson’s statement as to which gauge he used to measure the footballs pre-game?
• jddohio
Posted Jul 8, 2015 at 9:07 PM | Permalink
JonP “JD you are totally disregarding the video evidence showing McNally did exactly that, twice, in the same game. Who are we to believe video evidence or the refs CYA statements.”
You have a reasonable point here. I don’t know the answer. However, I do know that the officials said that McNally was not permitted to just walk out of the officials room prior to the game with the footballs. The most logical explanation is that, from the vantage point of the officials and the league, the later situation was viewed as different.
However, I still go back to McNally’s original statement: he didn’t know why he took the balls out of his own accord without permission or without being accompanied by the officials. I find this statement totally incredible. He could have said he had permission, he could have said he wanted to make sure he could wave to his mother-in-law, he could have said that it was a nice day outside. There was a reason he went out unaccompanied without permission, but when asked originally about it, he claimed to have no knowledge about something that happened in the very recent past.
Since the officials are only part-time workers and don’t work together full time, I believe it would be very difficult to get them all to lie together. On the other hand, the Patriot employees all work together and the Patriots have a history of cheating. I could be convinced otherwise, but from what I know of NFL officials, I believe they are far less likely to lie than Patriot employees.
JD
• chuckrr
Posted Jul 8, 2015 at 10:18 PM | Permalink
JD…Have you ever been asked a question that came out of nowhere? A question you never even considered? Now if he was guilty of this horrible crime he might tend to be a little prepared to answer the question. Criminals are good at that because they’ve already thought about what answers they’ll need to give when they are caught. On the other hand someone that doesn’t think they have done anything wrong might be tongue tied. Surprised that they are even asked the question.
His answer is totally believable if you think a little bit about human nature.
Why can’t you consider this bizarre possibility. None of them are lying. They just have a different perspective of what they think is reality.
• jddohio
Posted Jul 8, 2015 at 10:30 PM | Permalink
chuckrr “Have you ever been asked a question that came out of nowhere? A question you never even considered?”
This is not a question out of nowhere. This is a question that deals very specifically with McNally’s primary duties. I find it absolutely incredible that he would claim he had no idea why he took it upon himself to take the balls out without supervision or permission. Also, the fact that Brady loves deflated balls fills in more context.
JD
• chuckrr
Posted Jul 8, 2015 at 11:07 PM | Permalink
If the question wasn’t unexpected to him (out of nowhere) then he would have thought about it right? And the obvious answer is that he thought he had permission. Weather he was right or not is another question. It would make sense and be hard to argue that he wasn’t sincere since everyone else was leaving the locker room too. Your right in that his answer makes no sense unless he was startled by the question. And couldn’t understand why he was being asked the question.
Now here is another question for you. How did the Patriots deflate the balls on the road? It’s hard to believe Brady would like to use balls inflated to different pressures for different games. So how did they do it. McNally wasn’t there. What’s the context. Maybe there’s another joke Brady made to explain that
• MikeN
Posted Jul 18, 2015 at 1:49 PM | Permalink
Rules out recording verbatim responses.
• MikeN
Posted Jul 9, 2015 at 8:49 AM | Permalink
The Patriots ‘history of cheating’ has been covered elsewhere. Bill Belichick misunderstood a memo by the NFL. Another site went through the rules and showed that in fact the NFL’s memo is not consistent with the rest of the rulebook and Bill was right, though of course the recent memo would take precedence.
• MikeN
Posted Jul 9, 2015 at 10:36 AM | Permalink
>McNally’s original statement: he didn’t know why he took the balls out of his own accord without permission or without being accompanied by the officials. I find this statement totally incredible. He could have said …he claimed to have no knowledge about something that happened in the very recent past.
No, he said he was uncertain, and this is a third or fourth hand quote.
• Steve McIntyre
Posted Jul 9, 2015 at 1:51 PM | Permalink
McNally’s original statement: he didn’t know why he took the balls out of his own accord without permission or without being accompanied by the officials. I find this statement totally incredible. He could have said …he claimed to have no knowledge about something that happened in the very recent past.
As I mentioned in an earlier comment, this is a very inaccurate characterization of events. McNally had already received permission to move the balls to the waiting area and was waiting with the officials for the NFC championship game to end. Because they were waiting for another game to end, the usual sequence wasn’t followed in any event. McNally’s account, accoording to Wells, was that, when the NFC game ended, an NFL official said something like “we’re back on again” with the room quickly emptying. After hearing from an official that they were “on”, McNally picked up the balls and walked out of the Officials Locker Room towards the field, where he arrived a couple of minutes later, after his bathroom stop.
Based on other information, McNally said that he frequently took the balls unaccompanied, a claim supported by security guards but disputed by the NFL. IT’s possible that the security guards are lying, but it’s also possible that the NFL officials are CYA’ing. Based on the information that I’ve seen, I put more credence on the security guards, but I understand that reasonable people can disagree on this.
• jddohio
Posted Jul 9, 2015 at 11:58 PM | Permalink
Response to Steve Mc: I have made very clear that the dates of the statements made by McNally are very important to me. Your summary ignores the dates.
The quotes that support my statement was made on Jan. 18 & 21. The quote that supports your statement was made on 2/21.
The Jan. 18 statement:
“According to a report of the interview with McNally on the night of January 18, McNally told NFL Security representatives that he “decided to walk the balls out to the field,” and was “NOT CERTAIN WHY [HE] CHOSE TO GO OUT TO THE FIELD AT THIS TIME OR WITHOUT AN ESCORT.” [All caps by JD] McNally also told NFL Security during this interview that he walked directly to the field and that nothing unusual occurred during the walk from the locker room to the field.” [p.58]
********
Then 3 days later on Jan. 21: ” According to the interview report from an NFL Security interview of McNally on January 21, McNally said that he DID NOT KNOW WHY HE WOULD LEAVE THE LOCKER ROOM [upper case added by JD] with the game balls without being accompanied by game officials, and “just decided to leave the locker room at that time to go to the field.” (p. 59)
Then having had a month to think it over and talk with lawyers and Patriot officials he makes the statements you quote. “On February 12, 2015, we interviewed McNally on these topics as well. He explained to us that he told the game officials that he was moving the game balls to the sitting
room, where he watched the end of the NFC Championship Game for up to ten minutes. He estimated that there were twenty people in the sitting room at the time. According to McNally, when the NFC Championship Game ended shortly after the start of the overtime period, an unidentified NFL official said something like “we‟re back on again,” so he picked up the balls and began to walk out of the Officials Locker Room.”
Sorry but I am not much impressed by his 2/21 explanation which conflicts with his 2 earlier statements and which was made after he had a long time to talk to Patriot officials and lawyers and try to concoct a believable story. As I have said before, I believe McNally hours after he had gone on the field unaccompanied had to have known why he did. The fact that he claims not to know why is very damaging in my view.
JD
• jddohio
Posted Jul 10, 2015 at 1:04 AM | Permalink
Another potential problem with the 1/18 statement by McNally is that it can be read as not revealing that he stopped in a bathroom on the way to the field. I hadn’t noticed this until Steve made me look at the quotes more closely. McNally stated: “that he walked directly to the field”, which can be interpreted as not involving a bathroom stop. I am bringing this up for further discussion and context because I haven’t had the chance to look more closely at the context and someone else may be more familiar than I.
JD
• mpainter'm
Posted Jul 10, 2015 at 8:12 AM | Permalink
JD:
There is a lot of confusion on the requirement that the ball attendant (McNally) be escorted. In fact, there is no such requirement.
From the NFL protocol dealing with the football’s:
Rule 2: “….and the balls shall remain under the supervision of the referee until they are delivered to the ball attendant just prior to the start of the game”
It says nothing about escorting the ball attendant onto the playing field.
McNally’s inconsistencies are inconsequential and should not be magnified into a hangman’s noose.
To me the real issue is why Ted Wells would do this. He is an attorney and no doubt is familiar with the protocols.
• MikeN
Posted Jul 10, 2015 at 10:21 AM | Permalink
JD, what is your source for these quotes? The Wells Report doesn’t have them except as third or fourth hand, but you have them in quotation marks.
• Carrick
Posted Jul 10, 2015 at 2:57 PM | Permalink
mpainter:
Rule 2: “….and the balls shall remain under the supervision of the referee until they are delivered to the ball attendant just prior to the start of the game”
It says nothing about escorting the ball attendant onto the playing field.
Clearly the intent of Rule 2 is to prevent one of the teams from gaining a undue competitive advantage by modifying them after they’ve been tested:
If you argue that this rule is to be interpreted so as to allow the footballs to be delivered to the ball attendant so that he can disappear from sight and potentially modify the foot balls,then you are arguing for an interpretation that allows the home team to gain an undue competitive advantage here.
No football team is going to sign on to an interpretation that permits their competitor from gaining an undue advantage in away football games.
MikeN: These quotes have the quotation marks in the text too. See for example page 58.
• JD Ohio
Posted Jul 10, 2015 at 3:32 PM | Permalink
MikeN: What is the source for these quotes.
The investigations. Do you think that the investigators talked to McNally and did nothing to document their talks? The Patriots attorneys in their annotation website don’t dispute that McNally made the statements as reported in the Wells Report. Attorneys were undoubtedly involved in this in the beginning and they all know that dicussions with witnesses have to be well documented. The questioners did not just ask the questions and do nothing to document the questions and answers.
That being said it would be much preferable to see transcripts of the interviews. In particular, I would like to see the precise question asked when McNally originally said he walked directly to the field.
JD
• MikeN
Posted Jul 10, 2015 at 5:41 PM | Permalink
JD, I missed the quotes in the report, I see them now. They are no longer third or fourth hand, but second hand or third hand quotes since based on the ‘he’ I think we have to say that it is a summary of McNally’s answer by the investigators rather than a direct quote.
• MikeN
Posted Jul 10, 2015 at 5:49 PM | Permalink
Carrick, that’s not the only rule in the book. The teams are not allowed to tamper with the footballs.
The idea that this rule must cover every possibility is absurd. They might change the rule in light of events or issue a memo, but as written the rule does not cover that.
For example, based on what happened, we know that the referee wasn’t watching the footballs in the locker room either. Therefore McNally must have broken the rules by being in a position where he could take the footballs without the referee noticing, right?
• MikeN
Posted Jul 10, 2015 at 10:19 PM | Permalink
For example, at halftime they let McNally again take the balls by himself unaccompanied. So I guess he is again in violation.
The referees didn’t seem to think much of it. However, they told the Pats official they had better not deflate the footballs again.
• Carrick
Posted Jul 10, 2015 at 11:23 PM | Permalink
MikeN, the rulebook doesn’t have to cover every possibility because people generally understand the intent.
I’d say the remedy when you have a team representative breaking rules is to ban them from participating, which is pretty much where McNally is (I’m pretty sure the suspension/ban came from the NFL rather than the Patriots), and fine the team for which they were a representative. In a sport like this, generally you can’t impair his motion, especially if you’re the NFL officiating crew, rather than the Patriots security personnel.
• chuckrr
Posted Jul 11, 2015 at 7:10 AM | Permalink
The “intent” of the rule is irrelevant. How. the rule is enforced becomes the the way the rule perceived.. It’s obvious that this rule was lackadaisically enforced. Therefore it was understood to be not that important. So you ban somebody because now you arbitrarily decide to enforce a vaguely understood rule? Remember McNally thought he had permission to take the balls to the field…unsupervised…as he had done before. And as it says in the rule he can do. A rule becomes important to strictly enforce when it becomes news. So now this will be an important rule to enforce. Even though it’s obvious McNally did nothing else wrong ‘ It’s news . So now the “intent” will be scrutinized and the rule enforcement procedures analized. And McNally will be cleaning the bus treads off his jacket
• jddohio
Posted Jul 15, 2015 at 11:15 PM | Permalink
There has been enough time for everyone to review McNally’s first two statements (1/18 and 1/21) where he essentially stated that he did not know why he left the locker room without permission or an escort. On Feb. 12 he came up with a different explanation (sorry for earlier incorrect tranposition to 2/21)
Does anyone care to explain how McNally could TWICE close to the event claim he didn’t know why he went out to the field without permission or an escort, but would somehow remember on on a 2/12 third interview why he went out.
I see no reason to believe McNally’s third statement. I still find his first two statements incredible — he claims not to remember a simple detail that was very important to his work, within hours in one case (the 1/18 interview)after he performed his duties.
It is also interesting to note that in 2004 (p. 44) McNally escaped punishment for the introduction of practice balls into a game when he stated that he didn’t know how it occurred. The NFL stated that ““the Patriots have not provided a reasonable explanation for this incident.”
JD
• chuckrr
Posted Jul 16, 2015 at 2:21 PM | Permalink
JD,
Is there somewhere that you have seen the actual transcripts of the interviews? Because I haven’t. ..I’ve only seen a few paraphrased references to the conversations. It might be interesting to have some context…don’t you think? Even so the fact that he doesn’t give the answer you think would make sense might be a reflection on your assumptions. I have yet to see any provable lies in his statements.
• MikeN
Posted Jul 16, 2015 at 3:54 PM | Permalink
JD, we are not looking at direct quotes, or I think even statements taken directly from notes. My understanding is that the interview reports are summaries made from notes taken by the interragator(who writes the summary). This is what the FBI does, and one of the interrogators is a former FBI agent, and the other is a detective. Reading what’s in the Wells Report, the use of (he) suggests that it is not a direct quote from McNally.
What Chuck and I are suggesting is something like,
McNally feels he did nothing unusual.
Q:”Why did you go out to the field by yourself?”
Mc:huh?
Q:”Why didn’t you wait till you were instructed by Anderson?”
Mc: I’m not sure.
Also the Wells Report doesn’t say McNally said ‘I don’t know’ but rather ‘not certain’ is in quotes, which would again be from the summary written from notes.
I don’t know is from the second interview and not in quotes.
If he was being cagy about it, then your point is valid, but we don’t know that. Indeed, Wells reports this about Jastremski regarding the 50000 football, yet here nothing is said. I think it is more probable than not that a detective and an FBI agent would write if they thought defendant was being suspicious, and would be able to detect it in McNally.
• jddohio
Posted Jul 16, 2015 at 5:22 PM | Permalink
Mike N ” My understanding is that the interview reports are summaries made from notes taken by the interragator(who writes the summary). This is what the FBI does, and one of the interrogators is a former FBI agent, and the other is a detective. Reading what’s in the Wells Report, the use of (he) suggests that it is not a direct quote from McNally.”
It is theoretically possible that the quotes provided are from summaries. However, it would be grossly negligent not to have an audio tape (or something similar from the interviews)to confirm the statements made. People routinely deny statements that others quote them as having made, so any modestly qualified investigator would avoid that problem. There is no point in having an investigation and quoting people if there is no way to confirm what statements were made. This is very well known in the legal profession.
At about the 8th annotation from the end the Patriot attorneys in the annotation report discuss permission to leave the dressing room and go to the locker room. (Note in my view the attorneys intentionally try to give the false impression that McNally got permission to go to the field in the passage when they talk about permission to go to the locker room but omit any discussion of permission to go to the field.) This is the only passage I saw that dealt with, to some minor extent, with what permission that McNally had to move the balls. In contrast to the text messages section, which is very detailed, this annotation section is very short. Also, it does not dispute in anyway (nor did any other portion of the report that I saw) the statements recorded in the investigation that McNally didn’t know why he went to the field when he did. The fact that the annotation went over the text messages very thoroughly, but essentially skipped over McNally’s original description of what happened strongly indicates that he was correctly quoted.
JD
• MikeN
Posted Jul 16, 2015 at 6:04 PM | Permalink
Later in the Wells Report, they said that McNally in February denied making the statements attributed to him in January. If there were audio recordings, or even a transcript, that point would have been mentioned. Instead they referred to separate interviews with the interviewers who confirmed the accuracy.
I agree with you the Patriots are being misleading. Particularly, the Wells Report I think has McNally sitting right by the door to the hallway, while the Pats say none of the officials objected as he took this bag past them.
I still think McNally if he wanted to do something had better opportunity in the locker room, and that he is physically incapable of doing what they say he did.
• Posted Jul 16, 2015 at 6:31 PM | Permalink
JD, wellsreportcontext annotations are on the Executive Summary only, so if the executive summary doesn’t specifically address the statements recorded in the investigation, then the Patriots context doesn’t address it. They did explain McNally’s original description of what happened in at least 1 other place:
“With no notice to Patriots management, League security actually began investigating during the second half of the game when they began questioning Patriots ball boys. Consistent with that, Mr. McNally described the focus of his first interview as being on the role of ball boys. It was accurate for him to have stated that nothing unusual happened during the walk from the locker room to the field, since, as he later explained, his bathroom stop was nothing unusual. When later asked why he did not use the urinals in the Officials’ Locker Room or the chain gang room, he fully explained why — and his reasons are supported by the report’s conclusions about how crowded the Officials’ Locker Room area was (pgs. 54-55). One can draw no adverse inferences from an attendant deciding not to use the crowded facilities. If the investigators had found a single witness who had seen Mr. McNally routinely using the urinals in the Officials’ Locker Room prior to other games when the officials were doing their final pre-game preparations, they would have put that in the report. Similarly, no one instructed Mr. McNally not to use the bathroom he used, which is on his direct route from the Officials’ Locker Room to the field.”
Here’s another:
“What Mr. McNally actually described was exactly what the report stated happened before the AFC Championship Game — that he gets permission from the game officials to remove the footballs from where they reside in the dressing room of the Officials’ Locker Room. As the report acknowledges (pg. 55), Mr. McNally received precisely that permission: “Anderson also recalls that Mr. McNally, with Anderson’s permission, had moved the bags of footballs from the dressing room area towards the sitting room shortly after the officials returned from the player’s walk-through.” Thus, Mr. McNally had the referee’s permission to remove the footballs from the part of the dressing room where game officials congregate pre-game. He sat with the footballs in the sitting room and then, when the NFC Championship Game that everyone was watching in that sitting room ended, he took the footballs from the sitting room and out into the hallway in full view of numerous League and game officials. Even after halftime, when psi measurements had become an issue, Mr. McNally is seen on the security tape walking the footballs back to the field totally unaccompanied by any game or League official, but obviously with their full knowledge — even more than “general awareness” — that he was doing so. Again, no one told him to wait, to stop, or that he was doing anything wrong in taking the footballs from the Officials’ Locker Room to the field.”
I would ask you, If Anderson and other League Officials have never seen a ball attendant take the balls to the field unaccompanied in 19 years, AND they were all made aware of the accusations the Patriots may be tampering with footballs, can you explain why they would not find the “lost balls” re-test them, or use back up balls?
And after their panic pre-game, why would they again allow McNally to walk the balls out unattended at the end of half-time?
So it’s never happened in Anderson’s 19 years – it happens – the balls are measured at half time and because no one understands the ideal gas law it looks like there is clear evidence of tampering – annnndddddddd – Anderson lets McNally take the balls out to the field unaccompanied AGAIN. And neither says or does anything about it, AGAIN.
• chuckrr
Posted Jul 16, 2015 at 9:03 PM | Permalink
I’ll restate my point. I see no exact quotes only the Wells report paraphrase and interpretation of the interviews. As Dave says it reads like a prosecutorial argument. JD…You may place some trust that these people were trying to be impartial but I can’t. I see to many red flags We’ve seen plenty of these cases of prosecutorial overreach in the news recently. And I realize that my views are now biased by my opinion of the report. The point is that absent a direct provable lie your perspective on the interviews can be dramatically different, , which is influenced by our assumptions and biases. The only facts and provable information we have is what Steve has presented.
• jddohio
Posted Jul 16, 2015 at 10:04 PM | Permalink
Mike N “Later in the Wells Report, they said that McNally in February denied making the statements attributed to him in January.”
I never saw where McNally denied making the January statements. He just changed his story in February and came up with reasons he couldn’t state in January. If you can see where there is a place where McNally denied stating what was quoted for 1/18 & 1/19 please point it out. There is a big distinction here.
JD
• jddohio
Posted Jul 16, 2015 at 10:09 PM | Permalink
Chuck rrr “I’ll restate my point. I see no exact quotes only the Wells report paraphrase and interpretation of the interviews. As Dave says it reads like a prosecutorial argument. JD…You may place some trust that these people were trying to be impartial but I can’t.”
I don’t claim that the investigators are impartial, and if fact I don’t think they are impartial. I claim that in about 95% of the cases, they would realize that it is in their own best interest to be able to replicate what McNally said. Any competent investigator wouldn’t ask others to believe him, if he had a good way (audiotapes for instance) to back up his findings. Therefore, to help their clients investigators should have substantive ways of verifying their findings.
JD
• jddohio
Posted Jul 16, 2015 at 10:17 PM | Permalink
DaveM “I would ask you, If Anderson and other League Officials have never seen a ball attendant take the balls to the field unaccompanied in 19 years, AND they were all made aware of the accusations the Patriots may be tampering with footballs, can you explain why they would not find the “lost balls” re-test them, or use back up balls?
And after their panic pre-game, why would they again allow McNally to walk the balls out unattended at the end of half-time?
So it’s never happened in Anderson’s 19 years – it happens – the balls are measured at half time and because no one understands the ideal gas law it looks like there is clear evidence of tampering – annnndddddddd – Anderson lets McNally take the balls out to the field unaccompanied AGAIN. And neither says or does anything about it, AGAIN.”
Your basic issue is how did the officials screw up. In my view the answer is simple. They had the complex and very difficult job of refereeing a football game that would be seen in front of millions of people. That was their first priority. They were not trained to scientifically check the psi of footballs. They screwed up a reasonably complex matter that they didn’t do repeatedly in the midst of performing a very high pressure and complex job.
I would contrast that with McNally whose job could have been performed by a 14-year-old, and who couldn’t explain in January why he went out to the field without an escort or without gaining permission.
JD
• MikeN
Posted Jul 17, 2015 at 7:27 AM | Permalink
Page 96.
• MikeN
Posted Jul 17, 2015 at 7:30 AM | Permalink
Just as they had pressures pre-game and with overtime of the other game things became hectic. After the fact, they claim things differently.
I suspect if Wells had noticed and asked the referees about the halftime taking of the footballs, they would have said he’s not supposed to do that either.
• Posted Jul 17, 2015 at 4:43 PM | Permalink
JD “Your basic issue is how did the officials screw up. In my view the answer is simple. They had the complex and very difficult job of refereeing a football game that would be seen in front of millions of people. That was their first priority. They were not trained to scientifically check the psi of footballs. They screwed up a reasonably complex matter that they didn’t do repeatedly in the midst of performing a very high pressure and complex job.
I would contrast that with McNally whose job could have been performed by a 14-year-old, and who couldn’t explain in January why he went out to the field without an escort or without gaining permission.”
My point is not that the officials screwed up – I am saying their actions (or lack thereof) after they screwed up (when they discover the balls were walked to the field unattended) don’t square with how the wells report characterizes the situation.
I am trying to square who is more likely being truthful about whether McNally walking the balls to the field unattended was, 1. “never done before in 19 years” (Anderson and other refs Wells says he talked to), or 2. was done 50% of the time by McNally (McNally and 2 security personnel). Because I am saying that i believe walking unattended was normal for McNally, so when he’s initially being questioned about it, it’s insignificant to him, and his answers are given in that context.
The Wells report makes a big deal about how stressed out Anderson and Farley are about losing site of the balls. Wells also says he talked to numerous refs who had never seen this before, and Anderson has never seen it in his 19 years. The implication is that McNally was doing something nefarious by walking the balls unattended. Wells uses that, combined with him going into the bathroom to paint a picture that McNally was secretly carrying out a plan to deflate balls, as opposed to what McNally asserts – that he walks to the field unattended often, and uses the bathroom on the way at times. As hectic and complex as Anderson and Farley’s jobs may be, they had been alerted to be on the lookout for Pats ball tampering. If this was such a huge breach in protocol as Anderson, Farley and the Wells report asserts it is, then it’s inconceivable that they didn’t do something to correct this huge breach in protocol.
I am saying that Anderson’s/Wells assertions about balls never going to the field unattended don’t line up with Anderson’s/Farley’s actions after they discovered McNally walked the balls unattended, and therefore i don’t believe Anderson or the Wells report when they discount McNally’s assertion that he walks balls to the field unattended 50% of the time.
• Posted Jul 17, 2015 at 4:57 PM | Permalink
JD, “They were not trained to scientifically check the psi of footballs. They screwed up a reasonably complex matter that they didn’t do repeatedly in the midst of performing a very high pressure and complex job.”
I wouldn’t even say they screwed up, certainly not over checking psi of the footballs. I am saying that based on their testimony about McNally walking the balls unattended and how unprecedented it was, combined with them being alerted to be aware of tampering prior to the game, they should have taken corrective action based on their statements (they had never seen it happen before).
• JD Ohio
Posted Jul 18, 2015 at 9:00 AM | Permalink
MN Page 96
Your reference to p. 96 indicates that the NFL may not have adequately backed up the record of the January interviews. If that is the case, the NFL is so stupid that it deserves whatever it gets.
JD
• MikeN
Posted Jul 18, 2015 at 1:39 PM | Permalink
JD, They’ve adequately backed up the record, it is the record itself that is the problem.
My source is admittedly weak- a fiction book- but I was struck by how it matched the Wells Report, and I think may indeed be a common practice. The book was set in the past though.
• MikeN
Posted Jul 18, 2015 at 1:50 PM | Permalink
Rules out recording verbatim responses.
• MikeN
Posted Jul 18, 2015 at 1:51 PM | Permalink
The NFL is not getting anything, McNally is. Incidentally that was how the book described it with the FBI able to fabricate lying to law enforcement claims.
• Posted Jul 16, 2015 at 3:55 PM | Permalink
IMO the wells report reads like a prosecutorial advocacy paper, not an independent report, so I don’t concede that statements and descriptions of events in the report are fact. Much of it is written in a way that minimizes NFL issues, issues with the completeness of the evidence, and makes big, important inferences in ways that consistently don’t favor the Patriots. It reads like what a prosecutor might put together.
Re: McNally – His first interview was that night, 1/18 – He claims the focus was on the ball boys. His 2nd interview was 1/19. His 3rd interview was 1/21, where he claims investigators were demeaning, aggressive and accused him of lying. His 4th interview (the 1st with Wells) was in February.
Just because the explanations of things are odd or inconsistent in small details, doesn’t mean that they are lies IMO. The truth is often odd and people’s recollection of events that they thought were inconsequential at the time are usually not the Gods-eye truth IMO.
I believe the tone of the interviews changed during the 3rd interview, where it becomes clear to McNally that NFL thinks he tampered with footballs and that he is lying – so his answers in 1st and 2nd interview are in the context of him not being suspected of anything – and him trying to recollect events that were inconsequential to him at time they occurred vs. 3rd and 4th interview where he knows they think he’s lying.
• MikeN
Posted Jul 17, 2015 at 7:42 PM | Permalink
I think that’s what the investigators should do if they want to get to the truth.
For example, if they could convincingly say there is a security camera in the bathroom, we are getting the video now. One look at his face would have told them if he were innocent or guilty.
18. Posted Jul 8, 2015 at 6:32 PM | Permalink
JD, Agreed, if Wells interviewed all refs and all locker room attendants, then we would have a clear answer.
I think I’ve confused the issue – The first protocol is the referee gives his permission to take the balls from the officials locker room. I misunderstood the protocol to also say that the balls need to be walked to the field in the presence of an NFL official.
According to the wells report, the attendant EITHER needs permission OR needs to be accompanied to the field by an Official:
Here’s the quote: “No official could recall another time that McNally had removed game balls from the Officials
Locker Room and taken them to the field without either receiving permission from a game official or being accompanied by one or more game officials.”
Another quote from Anderson: “the footballs do not leave the locker room until the officials give express permission for them to
be brought to the field at or near the time the officials also walk to the field”
It doesn’t say they have to be brought to the field accompanied by an official.
So the actions that are in question that point to guilt is he left the room without express permission.
Walking the footballs to the field without an NFL official is fine, as long as you get permission from the ref in the Officials room.
19. Posted Jul 8, 2015 at 6:55 PM | Permalink
JD, I guess I’m disputing “the officials statements that the balls are to be taken to the field with officials present” I don’t read the wells report that way, as per my last post.
As further proof that my interpretation is correct – if this was the first time that any officials recall McNally walked the balls to the field without permission and without officials present, how was he deflating the balls in all of the Patriots other 8 home games of 2014?
Because if the deflator term in 1 text from May 2014 (offseason between 2013 and 2014 seasons) is referring to deflating footballs, then that means McNally was deflating footballs during the 2013 season and all of 2014 season.
As further anticdotal evidence against tampering:
If McNally was deflating balls for the entire 2014 season, how do you explain the texts in 2014 after the Jets game about the “refs F’d us, some of the balls are 16 psi”. You would think there’d be something in the texts about McNally f’ing up his deflating process. Not about the refs screwing them. And he didn’t deflate balls during the jets game if some were 16 psi.
That’s the thing – when you put all the “evidence” of guilt together, the information doesn’t make sense.
• MikeN
Posted Jul 8, 2015 at 7:44 PM | Permalink
The Wells Report said he could deflate in the locker room during other games.
• jddohio
Posted Jul 8, 2015 at 9:12 PM | Permalink
D 1964 “If McNally was deflating balls for the entire 2014 season …” I am not claiming that and I don’t believe that the Wells report is necessarily claiming that. The point of bringing up the “deflator” issue and the texts surrounding it was to show that the issue of ball inflation was very important to Brady. If it was important to Brady, it would have to be important to McNally, which shows a motivation for deflating balls and at one point of time. If these texts weren’t hightlighted, people would be asking what was McNally’s motivation. All the league is trying to do is show one instance of “illegal” deflation.
JD
• chuckrr
Posted Jul 8, 2015 at 10:03 PM | Permalink
Do you have any evidence that Brady wanted the balls inflated below 12.5 psi? Anything? That would be motivation. The only motivation that I see is Brady didn’t want them inflated to 16 psi and preferred 12.5
• jddohio
Posted Jul 8, 2015 at 10:25 PM | Permalink
chuck: “Do you have any evidence that Brady wanted the balls inflated below 12.5 psi? Anything? That would be motivation.”
He loved deflated balls. That is all I need to know. You may not agree. Just because someone is not caught specifically admitting something does not mean that it is unreasonable to reach a conclusion regarding prohibited behavior. Ftn. 15 of the report: “Brady made public statements concerning his preference for a “deflated” ball at least as early as 2011. Specifically, during a November 14, 2011 interview on Boston‟s WEEI radio, Brady praised Patriots tight end Rob Gronkowski for powerfully spiking footballs after scoring touchdowns because of its impact on the ball.
Brady stated that “I love that, because I like the deflated ball.”
JD
• chuckrr
Posted Jul 8, 2015 at 10:30 PM | Permalink
Well all I can say to that is that the human mind is an amazing and terrifying thing.
• chuckrr
Posted Jul 8, 2015 at 10:34 PM | Permalink
Oh…and your explanation is a good example of why you want to avoid the legal system
• MikeN
Posted Jul 9, 2015 at 7:16 AM | Permalink
They have to be claiming that he did it before, if they are going to bring up a text from the previous season. Unless they are arguing in May they hatched a plan to inflate only in the AFC championship game.
• Posted Jul 9, 2015 at 4:00 PM | Permalink
JD, re: wells report claiming McNally was deflating balls in 2013 and 2014 seasons… I believe they do claim that. But I guess they aren’t necessarily claiming it was every game.
1. The wells report cites McNally calling himself “the deflator” in a May 9 2014 text as a key piece of evidence in proving that McNally had deflated footballs prior to the AFC game. Although this is the only time McNally calls himself the Deflator, wells references this nickname 16 times in the report to support their arguments. They clearly are claiming his calling himself the “deflator” in may 2014 is referring to illegally deflating footballs.
So if we are to accept the wells interpretation of this text (used in the offseason between the 2013 and 2014 seasons), i think you have to conclude than McNally was deflating footballs at some point before this text.
2. Additionally, they point to a series of texts after the Jets 2014 game as evidence that McNally was deflating (where McNally is mad at Brady and saying he’s going to deliver balloons the next game). Here they claim the texts show that McNally was going to deflate balls in the next 2014 home game, because they reference Jastremski saying he can’t wait to give McNally the big needle.
I’m not sure how to reconcile the fact that the balls after the Jets home game are at 16 psi – Is the wells report claiming that McNally didn’t deflate the balls before the jets game? They point to the texts after the jets game as proof that McNally is upset that Brady was upset about the balls being overinflated. – so Brady must have been upset that McNally and Jastremski didn’t deflate them?
3. They then reference a Nov 2014 text during halftime of an away game where McNally sees Jastremski on TV and texts – “delflate and give someone that jacket”. They claim this is in regards to deflating illegally.
So based on Wells interpretation of 1, 2 and 3, Wells is claiming McNally was deflating balls illegally sometime in 2013 and in some games during 2014. He is also claiming Jastremski was deflating balls at some point prior to the Nov 30th 2014 Away game.
20. MikeN
Posted Jul 8, 2015 at 7:56 PM | Permalink
JD, you are emphasizing that McNally’s initial statement should be given more weight, and you discount his later statements. I’m curious, what is the later statement that McNally gave to explain his actions?
• jddohio
Posted Jul 8, 2015 at 9:17 PM | Permalink
MikeN See D 1964 at 3:33 where he summarizes McNally’s later statements. Here is part of it: ” McNally claimed that his actions on the day of the AFC Championship Game were not unusual. In his account, the game balls remainin the locker room until he believes it is time to take them to the field. According to McNally, he brings the game balls to the fieldwhen he deems fit. He said that he generally asks permission or alerts the officials before he moves the game balls from the dressing room to the sitting room, but does not ask or alert them again before leaving the locker room and taking the balls to the field.”
JD
• MikeN
Posted Jul 9, 2015 at 8:41 PM | Permalink
Tough to say without a transcript, but it feels like a different question is being answered. Not ‘why did you go by yourself’ but instead he is explaining what he didn’t do.
• Posted Jul 10, 2015 at 9:50 PM | Permalink
I see no issue with this statement. Am I to believe that a 300 pound man waddling past a room full of officially with two giant bags containing 24 footballs was missed by the referees? They saw this happenned and then found it so unusual they did nothing? He then went down the hallway by another official who found it so unusual he did nothing? Then after the big halftime measuring fiasco they were so concerned about this “unusual” behavior they finished measuring handed the balls to McNally where he again walked unaccompanied to th field past the officials.
It really isn’t difficult to see by the actions ON CAMERA that McNally is telling the truth about his routine, and the refs can’t even justify their own actions in that very game… Twice… On alert for anything unusual.
• MikeN
Posted Jul 10, 2015 at 10:18 PM | Permalink
My reading of the Wells Report is that McNally would have been very close to the door when he left the locker room.
21. jddohio
Posted Jul 8, 2015 at 10:46 PM | Permalink
Also, those of you relying on the wellsreportcontext.com, should understand that it was apparently prepared by the Patriots attorney who has a duty to paint the facts in a manner designed to support the Patriots. http://www.msn.com/en-us/sports/nfl/the-patriots-made-a-truther-website-about-the-wells-report/ar-BBjMQ3N?ocid=ansUSAsports11
I say apparently because I can’t find the article’s reference to an annotation by the Patriots attorney Goldberg. As it stands now, I don’t see the source of who is responsible for the website. Maybe others can find it. I believe that if the site is to be credible, its authorship should be disclosed and there is a very good chance the site was created by the Patriots attorney.
JD
Steve: JD, you’re being uncharacteristically inaccurate in this discussion. The website http://wellsreportcontext.com/ explicitly identified Daniel Goldberg, attorney for the Patriots as author in the second paragraph of its home page as follows:
These points, and others, are addressed in greater detail in the following Annotations to the Executive Summary of the Wells Report by Daniel L. Goldberg, a senior partner in the Boston office of Morgan Lewis and who represented the Patriots and was present during all of the interviews of Patriots personnel conducted at Gillette Stadium. Our intention is to provide additional context for balance and consideration.
In contrast, Exponent’s report did not identify any of the authors.
• chuckrr
Posted Jul 8, 2015 at 11:14 PM | Permalink
Are you assuming the Wells Report is an impartial unbiased independent study of the events?
• jddohio
Posted Jul 8, 2015 at 11:27 PM | Permalink
No, I believe it is also biased. Personally, I give little credence to either report per se. It is the events revealed in the reports that matter to me. I have repeated them multiple times and will not repeat again what I have considered to be important.
JD
• jddohio
Posted Jul 9, 2015 at 12:02 AM | Permalink
Steve: “JD, you’re being uncharacteristically inaccurate in this discussion.” Thanks for pointing out my error. It is late at night and I missed it. I was looking for annotations at the top or bottom of the page where attorneys would typically put them.
JD
22. MikeN
Posted Jul 9, 2015 at 7:43 AM | Permalink
The Wells Report does not declare that Jastremski appeared evasive or dishonest when answering questions about deflation. They only say it with regards to a 50,000 yard football that Tom Brady signed. Jastremski has either lied to his family members about this, or has secretly taken a football that he should not have with the Hall of Fame having the wrong ball. So there is evidence that the Wells investigative team were able to detect lies told by Jastremski, yet they did not detect anything with regards to deflation. Wells Report chose not to emphasize this lack of evasive tells in their analysis.
23. MikeN
Posted Jul 9, 2015 at 7:59 AM | Permalink
Veteran FBI agent and detective interviewed McNally the night of the game, and no mention is made that they detected any dishonesty or evasiveness on the part of McNally in the Wells Report, only that McNally later denied the statements attributed to him and what was discussed in the interview.
24. Jon P
Posted Jul 9, 2015 at 8:08 AM | Permalink
JD,
Why did Wells not take Anderson at his word for which gauge he used to measure the footballs pre-game?
Also, are you related to Nick Stokes?
• MikeN
Posted Jul 9, 2015 at 8:41 AM | Permalink
Not fair. JD’s been quite reasonable though late to the thread.
• Jon P
Posted Jul 9, 2015 at 9:16 AM | Permalink
Meant it in a light hearted way.
25. MikeN
Posted Jul 9, 2015 at 8:10 AM | Permalink
How does the Master Gauge work?
There are four conversions given,
Logo 11.49 -> 11.21 and 12.74-> 12.40
Non-Logo 11.11 ->11.09 and 12.33->12.29
26. MikeN
Posted Jul 9, 2015 at 8:21 AM | Permalink
Exponent had some people repeat the process of deflation into a bathroom. None of the three experimenters managed to deflate by less than .6 psi for a single football.
They also neglected to include a second bag of footballs in their experiment, but since all of them got it done in 70 seconds, this is not an issue.
• Carrick
Posted Jul 9, 2015 at 9:27 AM | Permalink
You can adjust how much it’s deflated by, by using different gauges of the needle, as well as by how long you let it deflate.
• Carrick
Posted Jul 9, 2015 at 9:30 AM | Permalink
This is also why the discussion of different needle sizes between Jastremski and McNally is so interesting.
27. MikeN
Posted Jul 10, 2015 at 10:53 AM | Permalink
So is the Exponent analysis and the text messages analogous to Tiljander and bristlecones?
28. joe
Posted Jul 12, 2015 at 2:48 PM | Permalink
Question?
What is the natural rate of deflation of a football that is idle?
What is the natural rate of deflation of a football used during a game?
I dont recall any comment in the wells report as to the natural rate of deflation.
As a point of reference, my bicycle tires will be inflated to 115psi before the start of every ride, 24 hours later, the typical pressure will be in the range of 95psi-100psi – effectively a 12-17% drop in psi over a 24 hour period. Assuming the rate of deflation of a football is comparable to my bicycle tires, then approx 0.12psi-0.2psi drop would be expected over the 3.0 hours (1.5 hours before the start of the game through halftime plus the 1.5 hours before game time at initial inflation of the balls)
• Jeff Westcott
Posted Jul 15, 2015 at 9:16 AM | Permalink
The rubber bladder in a football is substantially thicker than your presta valve inner tube, and the starting pressure in your tire is an order of magnitude greater than the starting pressure in the ball. The tire leaks fast through the tube itself as the mechanical valve is very tight. The ball leaks primarily through the valve that is a rubber seal that can leak faster as there is less and less pressure creating the seal. Two very different arrangements.
• joe
Posted Jul 15, 2015 at 11:53 AM | Permalink
Jeff – I agree that the two are different – (bicycle tire v football). I would also note that the bicycle tire during a ride receives substantially more stress, but also has more heat build up during the ride with offsets the lost air during the ride so that the difference in PSI before and after the ride is very small. Most of the lost PSI starts after the ride.
My main point – is that there is a natural rate of deflation which may in fact be negligible, but the Wells report did not appear to address the issue and appears to have operated on the assumption that the rate was zero.
29. chuckrr
Posted Jul 12, 2015 at 5:20 PM | Permalink
Spalding says their basketballs loose half the air pressure at a constant rate over the course of a year. So probably about .002 lbs for 3 hours. But I doubt anyone has actually checked
• chuckrr
Posted Jul 12, 2015 at 8:53 PM | Permalink
lose not loose….anyway I don’t think it’s anything significant. I will say that basketballs even at 50 F are deflated very noticeably and it seems like we would have to put air more in the winter. So they may lose air more when they are more deflated. I know that doesn’t seem logical.
30. rogerknights
Posted Jul 13, 2015 at 8:30 PM | Permalink
The Patriots have had about 33% fewer fumbles than other teams over the past dozen years, I’ve read. And fewer dropped passes.
It makes for a better game when there are fewer of those. The outcome is less unfair. And there are fewer “goats.”
And it makes for a less violent game, because it’s less likely that a vicious tackle will dislodge a ball, so there is less incentive to do so. Therefore, it makes for fewer injuries.
Therefore football’s officialdom, at all levels, should decrease the PSI standard by 10%.
31. Posted Jul 14, 2015 at 4:00 PM | Permalink
rogerknights, the info you read about pats fumbling 33% less than other teams has been debunked – that study excluded dome teams and included kick returns (where they use a standard kicking ball). Studies that look at the right statistics show the pats have been consistently at the top of the list on fewest fumbles, along with a few other teams. QB’s account for a significant portion of fumbles (holding on to the ball too long instead of throwing it away when no one is open, scrambling/running, etc.). Teams with qb’s who get rid of the ball quickly are at the top – peyton manning’s teams, brady, brees for NO and Ryan for Atl.
Here’s one analysis – http://www.backpicks.com/2015/05/17/fumbling-statistics-and-patriot-trends/
32. Posted Jul 16, 2015 at 6:53 PM | Permalink
Has anyone seen the argument that the logo gauge had to be used for pats pre-game check because:
1. Pats gauge was used by Pats to set balls at 12.6. 2. Anderson gauged Pats balls at 12.5. Therefore pats gauge and Anderson’s gauge must be calibrated the same.
2. Pats gauge was used to measure the ball the colts intercepted in 1st half – the avg. of the 3 separate readings taken was 11.52, and the 11.52 avg. matches the average of the half time logo readings (avg of logo half time readings = 11.49)
Or conversely, If non-logo gauge was used pre-game, then the intercepted ball measured with pats gauge should match the non-logo half time readings. non-logo avg readings were 11.11.
Here is the detail from Wells report:
From page 70: “The pressure of the Patriots ball that had been intercepted by the Colts was separately tested three times and the measurements—11.45, 11.35 and 11.75 psi, respectively– were written on athletic tape that had been placed on the ball for identification.”
Footnote from page 65. “The ball from the D’Qwell Jackson interception was tested by a gauge that Wells believes was the Patriots’ regular gauge”
Am I missing something here? Anderson’s pre-game gauge readings matches pats gauge readings, pats gauge reads intercepted ball at avg. of 11.52, logo half time readings avg. 11.49. Therefore Anderson used logo gauge pre-game.
If logo gauge was used pre-game, then pats balls show no evidence of tampering.
• MikeN
Posted Jul 16, 2015 at 9:49 PM | Permalink
If the measurements of the intercepted ball had been considered, the entire Exponent work would have needed to be thrown out due to margin of error.
33. MikeN
Posted Jul 18, 2015 at 5:51 PM | Permalink
Another argument I’ve seen is that the act of adding air to the football would increase the inside temperature of the football. Anderson inflated two footballs pregame. If they started at 12.0, then there would be a 1.5C increase in temperature. If you take out the two low numbers, you get an average of 11.22 instead of 11.11. | 52,506 | 235,893 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2023-23 | latest | en | 0.928138 |
https://zbmath.org/?q=an:0604.58015&format=complete | 1,726,290,534,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651548.18/warc/CC-MAIN-20240914025441-20240914055441-00657.warc.gz | 1,011,196,945 | 9,951 | ## On the Mountain Pass Lemma.(English)Zbl 0604.58015
Differential equations and their applications, Equadiff 6, Proc. 6th Int. Conf., Brno/Czech. 1985, Lect. Notes Math. 1192, 203-208 (1986).
[For the entire collection see Zbl 0595.00009.]
This is an expository paper on generalizations of the Mountain Pass Lemma (MPL). It extends the MPL in two aspects: (a) from a Banach space to a closed convex subset; (b) from the strong separation condition of values of a function to a weaker one. Theorem. Suppose that a $$C^ 1$$ function f satisfies the Palais-Smale condition with respect to a closed convex set C of a Banach space, and that $$\exists \alpha \in R^ 1$$ such that $Sup\{f(x)| \quad x\in \partial Q\}\leq \alpha,\quad Sup\{f(x)| \quad x\in Q\}<\infty,\quad f(x)>\alpha,\quad \forall x\in S,$ where Q and S are two closed subsets of C, $$\partial Q$$ and S link with respect to C. Then one of the three possibilities occurs; (1) $$\alpha$$ is an accumulation point of critical values. (2) $$\alpha$$ is a critical value with uncountable critical points. (3) $$c=\inf_{A\in F}\text{Sup}_{x\in A}f(x)$$ is a critical value, where $$F=\{A=\phi (Q)|\phi\in C(Q,C)$$, with $$\phi |_{\partial Q}=id|_{\partial Q}\}.$$
Three applications on multiple solutions of variational inequalities, semilinear elliptic BVP, and minimal surfaces are presented. The first includes some new results and for the last two refer the author [Sci. Sin., Ser. A 26, 1241-1255 (1983; Zbl 0544.35044); 1256-1265 (1983; Zbl 0544.35045)] and the author and J. Eells [Acta Math. Sin., New Ser. 2, 233-247 (1986)].
### MSC:
5.8e+06 Abstract critical point theory (Morse theory, Lyusternik-Shnirel’man theory, etc.) in infinite-dimensional spaces 5.8e+36 Variational inequalities (global problems) in infinite-dimensional spaces 5.8e+13 Variational problems concerning minimal surfaces (problems in two independent variables)
### Citations:
Zbl 0595.00009; Zbl 0544.35044; Zbl 0544.35045 | 605 | 1,969 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2024-38 | latest | en | 0.789465 |
http://commons.apache.org/proper/commons-math/javadocs/api-3.2/org/apache/commons/math3/analysis/interpolation/NevilleInterpolator.html | 1,397,780,038,000,000,000 | text/html | crawl-data/CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00378-ip-10-147-4-33.ec2.internal.warc.gz | 46,151,394 | 3,521 | ## org.apache.commons.math3.analysis.interpolation Class NevilleInterpolator
```java.lang.Object
org.apache.commons.math3.analysis.interpolation.NevilleInterpolator
```
All Implemented Interfaces:
Serializable, UnivariateInterpolator
`public class NevilleInterpolatorextends Objectimplements UnivariateInterpolator, Serializable`
Implements the Neville's Algorithm for interpolation of real univariate functions. For reference, see Introduction to Numerical Analysis, ISBN 038795452X, chapter 2.
The actual code of Neville's algorithm is in PolynomialFunctionLagrangeForm, this class provides an easy-to-use interface to it.
Since:
1.2
Version:
\$Id: NevilleInterpolator.java 1379904 2012-09-01 23:54:52Z erans \$
Serialized Form
Constructor Summary
`NevilleInterpolator()`
Method Summary
` PolynomialFunctionLagrangeForm` ```interpolate(double[] x, double[] y)```
Computes an interpolating function for the data set.
Methods inherited from class java.lang.Object
`clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait`
Constructor Detail
### NevilleInterpolator
`public NevilleInterpolator()`
Method Detail
### interpolate
```public PolynomialFunctionLagrangeForm interpolate(double[] x,
double[] y)
throws DimensionMismatchException,
NumberIsTooSmallException,
NonMonotonicSequenceException```
Computes an interpolating function for the data set.
Specified by:
`interpolate` in interface `UnivariateInterpolator`
Parameters:
`x` - Interpolating points.
`y` - Interpolating values.
Returns:
a function which interpolates the data set
Throws:
`DimensionMismatchException` - if the array lengths are different.
`NumberIsTooSmallException` - if the number of points is less than 2.
`NonMonotonicSequenceException` - if two abscissae have the same value. | 411 | 1,802 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2014-15 | longest | en | 0.498521 |
https://us.metamath.org/mpeuni/0ellimcdiv.html | 1,721,321,224,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514848.78/warc/CC-MAIN-20240718161144-20240718191144-00805.warc.gz | 506,126,166 | 27,437 | Mathbox for Glauco Siliprandi < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > Mathboxes > 0ellimcdiv Structured version Visualization version GIF version
Theorem 0ellimcdiv 42284
Description: If the numerator converges to 0 and the denominator converges to a nonzero number, then the fraction converges to 0. (Contributed by Glauco Siliprandi, 11-Dec-2019.)
Hypotheses
Ref Expression
0ellimcdiv.f 𝐹 = (𝑥𝐴𝐵)
0ellimcdiv.g 𝐺 = (𝑥𝐴𝐶)
0ellimcdiv.h 𝐻 = (𝑥𝐴 ↦ (𝐵 / 𝐶))
0ellimcdiv.b ((𝜑𝑥𝐴) → 𝐵 ∈ ℂ)
0ellimcdiv.c ((𝜑𝑥𝐴) → 𝐶 ∈ (ℂ ∖ {0}))
0ellimcdiv.0limf (𝜑 → 0 ∈ (𝐹 lim 𝐸))
0ellimcdiv.d (𝜑𝐷 ∈ (𝐺 lim 𝐸))
0ellimcdiv.dne0 (𝜑𝐷 ≠ 0)
Assertion
Ref Expression
0ellimcdiv (𝜑 → 0 ∈ (𝐻 lim 𝐸))
Distinct variable groups: 𝑥,𝐴 𝜑,𝑥
Allowed substitution hints: 𝐵(𝑥) 𝐶(𝑥) 𝐷(𝑥) 𝐸(𝑥) 𝐹(𝑥) 𝐺(𝑥) 𝐻(𝑥)
Proof of Theorem 0ellimcdiv
Dummy variables 𝑢 𝑣 𝑤 𝑦 𝑧 are mutually distinct and distinct from all other variables.
StepHypRef Expression
1 0cnd 10627 . 2 (𝜑 → 0 ∈ ℂ)
2 0ellimcdiv.d . . . . . . . . 9 (𝜑𝐷 ∈ (𝐺 lim 𝐸))
3 0ellimcdiv.c . . . . . . . . . . . 12 ((𝜑𝑥𝐴) → 𝐶 ∈ (ℂ ∖ {0}))
43eldifad 3896 . . . . . . . . . . 11 ((𝜑𝑥𝐴) → 𝐶 ∈ ℂ)
5 0ellimcdiv.g . . . . . . . . . . 11 𝐺 = (𝑥𝐴𝐶)
64, 5fmptd 6859 . . . . . . . . . 10 (𝜑𝐺:𝐴⟶ℂ)
7 0ellimcdiv.f . . . . . . . . . . 11 𝐹 = (𝑥𝐴𝐵)
8 0ellimcdiv.b . . . . . . . . . . 11 ((𝜑𝑥𝐴) → 𝐵 ∈ ℂ)
9 0ellimcdiv.0limf . . . . . . . . . . 11 (𝜑 → 0 ∈ (𝐹 lim 𝐸))
107, 8, 9limcmptdm 42270 . . . . . . . . . 10 (𝜑𝐴 ⊆ ℂ)
11 limcrcl 24481 . . . . . . . . . . . 12 (𝐷 ∈ (𝐺 lim 𝐸) → (𝐺:dom 𝐺⟶ℂ ∧ dom 𝐺 ⊆ ℂ ∧ 𝐸 ∈ ℂ))
122, 11syl 17 . . . . . . . . . . 11 (𝜑 → (𝐺:dom 𝐺⟶ℂ ∧ dom 𝐺 ⊆ ℂ ∧ 𝐸 ∈ ℂ))
1312simp3d 1141 . . . . . . . . . 10 (𝜑𝐸 ∈ ℂ)
146, 10, 13ellimc3 24486 . . . . . . . . 9 (𝜑 → (𝐷 ∈ (𝐺 lim 𝐸) ↔ (𝐷 ∈ ℂ ∧ ∀𝑦 ∈ ℝ+𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < 𝑦))))
152, 14mpbid 235 . . . . . . . 8 (𝜑 → (𝐷 ∈ ℂ ∧ ∀𝑦 ∈ ℝ+𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < 𝑦)))
1615simprd 499 . . . . . . 7 (𝜑 → ∀𝑦 ∈ ℝ+𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < 𝑦))
1715simpld 498 . . . . . . . . 9 (𝜑𝐷 ∈ ℂ)
18 0ellimcdiv.dne0 . . . . . . . . 9 (𝜑𝐷 ≠ 0)
1917, 18absrpcld 14804 . . . . . . . 8 (𝜑 → (abs‘𝐷) ∈ ℝ+)
2019rphalfcld 12435 . . . . . . 7 (𝜑 → ((abs‘𝐷) / 2) ∈ ℝ+)
21 breq2 5037 . . . . . . . . . 10 (𝑦 = ((abs‘𝐷) / 2) → ((abs‘((𝐺𝑣) − 𝐷)) < 𝑦 ↔ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)))
2221imbi2d 344 . . . . . . . . 9 (𝑦 = ((abs‘𝐷) / 2) → (((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < 𝑦) ↔ ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2))))
2322rexralbidv 3263 . . . . . . . 8 (𝑦 = ((abs‘𝐷) / 2) → (∃𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < 𝑦) ↔ ∃𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2))))
2423rspccva 3573 . . . . . . 7 ((∀𝑦 ∈ ℝ+𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < 𝑦) ∧ ((abs‘𝐷) / 2) ∈ ℝ+) → ∃𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)))
2516, 20, 24syl2anc 587 . . . . . 6 (𝜑 → ∃𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)))
26 simpl1l 1221 . . . . . . . . . 10 ((((𝜑𝑧 ∈ ℝ+) ∧ (𝑣𝐴 → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2))) ∧ 𝑣𝐴) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧)) → 𝜑)
27 simpl3 1190 . . . . . . . . . 10 ((((𝜑𝑧 ∈ ℝ+) ∧ (𝑣𝐴 → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2))) ∧ 𝑣𝐴) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧)) → 𝑣𝐴)
28 simpr 488 . . . . . . . . . . 11 ((((𝜑𝑧 ∈ ℝ+) ∧ (𝑣𝐴 → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2))) ∧ 𝑣𝐴) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧)) → (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧))
29 simpl2 1189 . . . . . . . . . . 11 ((((𝜑𝑧 ∈ ℝ+) ∧ (𝑣𝐴 → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2))) ∧ 𝑣𝐴) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧)) → (𝑣𝐴 → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2))))
3027, 28, 29mp2d 49 . . . . . . . . . 10 ((((𝜑𝑧 ∈ ℝ+) ∧ (𝑣𝐴 → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2))) ∧ 𝑣𝐴) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧)) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2))
3119rpcnd 12425 . . . . . . . . . . . . . . . 16 (𝜑 → (abs‘𝐷) ∈ ℂ)
32312halvesd 11875 . . . . . . . . . . . . . . 15 (𝜑 → (((abs‘𝐷) / 2) + ((abs‘𝐷) / 2)) = (abs‘𝐷))
3332eqcomd 2807 . . . . . . . . . . . . . 14 (𝜑 → (abs‘𝐷) = (((abs‘𝐷) / 2) + ((abs‘𝐷) / 2)))
3433oveq1d 7154 . . . . . . . . . . . . 13 (𝜑 → ((abs‘𝐷) − ((abs‘𝐷) / 2)) = ((((abs‘𝐷) / 2) + ((abs‘𝐷) / 2)) − ((abs‘𝐷) / 2)))
35 2cnd 11707 . . . . . . . . . . . . . . . 16 (𝜑 → 2 ∈ ℂ)
36 2ne0 11733 . . . . . . . . . . . . . . . . 17 2 ≠ 0
3736a1i 11 . . . . . . . . . . . . . . . 16 (𝜑 → 2 ≠ 0)
3817, 35, 37absdivd 14811 . . . . . . . . . . . . . . 15 (𝜑 → (abs‘(𝐷 / 2)) = ((abs‘𝐷) / (abs‘2)))
39 2re 11703 . . . . . . . . . . . . . . . . . 18 2 ∈ ℝ
4039a1i 11 . . . . . . . . . . . . . . . . 17 (𝜑 → 2 ∈ ℝ)
41 0le2 11731 . . . . . . . . . . . . . . . . . 18 0 ≤ 2
4241a1i 11 . . . . . . . . . . . . . . . . 17 (𝜑 → 0 ≤ 2)
4340, 42absidd 14778 . . . . . . . . . . . . . . . 16 (𝜑 → (abs‘2) = 2)
4443oveq2d 7155 . . . . . . . . . . . . . . 15 (𝜑 → ((abs‘𝐷) / (abs‘2)) = ((abs‘𝐷) / 2))
4538, 44eqtr2d 2837 . . . . . . . . . . . . . 14 (𝜑 → ((abs‘𝐷) / 2) = (abs‘(𝐷 / 2)))
4645oveq2d 7155 . . . . . . . . . . . . 13 (𝜑 → ((abs‘𝐷) − ((abs‘𝐷) / 2)) = ((abs‘𝐷) − (abs‘(𝐷 / 2))))
4720rpcnd 12425 . . . . . . . . . . . . . 14 (𝜑 → ((abs‘𝐷) / 2) ∈ ℂ)
4847, 47pncand 10991 . . . . . . . . . . . . 13 (𝜑 → ((((abs‘𝐷) / 2) + ((abs‘𝐷) / 2)) − ((abs‘𝐷) / 2)) = ((abs‘𝐷) / 2))
4934, 46, 483eqtr3rd 2845 . . . . . . . . . . . 12 (𝜑 → ((abs‘𝐷) / 2) = ((abs‘𝐷) − (abs‘(𝐷 / 2))))
50493ad2ant1 1130 . . . . . . . . . . 11 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → ((abs‘𝐷) / 2) = ((abs‘𝐷) − (abs‘(𝐷 / 2))))
5145eqcomd 2807 . . . . . . . . . . . . . 14 (𝜑 → (abs‘(𝐷 / 2)) = ((abs‘𝐷) / 2))
52513ad2ant1 1130 . . . . . . . . . . . . 13 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (abs‘(𝐷 / 2)) = ((abs‘𝐷) / 2))
5352oveq2d 7155 . . . . . . . . . . . 12 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → ((abs‘𝐷) − (abs‘(𝐷 / 2))) = ((abs‘𝐷) − ((abs‘𝐷) / 2)))
5417adantr 484 . . . . . . . . . . . . . . . 16 ((𝜑𝑣𝐴) → 𝐷 ∈ ℂ)
5554abscld 14792 . . . . . . . . . . . . . . 15 ((𝜑𝑣𝐴) → (abs‘𝐷) ∈ ℝ)
56553adant3 1129 . . . . . . . . . . . . . 14 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (abs‘𝐷) ∈ ℝ)
576ffvelrnda 6832 . . . . . . . . . . . . . . . . 17 ((𝜑𝑣𝐴) → (𝐺𝑣) ∈ ℂ)
58573adant3 1129 . . . . . . . . . . . . . . . 16 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (𝐺𝑣) ∈ ℂ)
5958abscld 14792 . . . . . . . . . . . . . . 15 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (abs‘(𝐺𝑣)) ∈ ℝ)
60173ad2ant1 1130 . . . . . . . . . . . . . . . . 17 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → 𝐷 ∈ ℂ)
6160, 58subcld 10990 . . . . . . . . . . . . . . . 16 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (𝐷 − (𝐺𝑣)) ∈ ℂ)
6261abscld 14792 . . . . . . . . . . . . . . 15 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (abs‘(𝐷 − (𝐺𝑣))) ∈ ℝ)
6359, 62readdcld 10663 . . . . . . . . . . . . . 14 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → ((abs‘(𝐺𝑣)) + (abs‘(𝐷 − (𝐺𝑣)))) ∈ ℝ)
6456rehalfcld 11876 . . . . . . . . . . . . . . 15 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → ((abs‘𝐷) / 2) ∈ ℝ)
6559, 64readdcld 10663 . . . . . . . . . . . . . 14 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → ((abs‘(𝐺𝑣)) + ((abs‘𝐷) / 2)) ∈ ℝ)
6657, 54pncan3d 10993 . . . . . . . . . . . . . . . . . 18 ((𝜑𝑣𝐴) → ((𝐺𝑣) + (𝐷 − (𝐺𝑣))) = 𝐷)
6766eqcomd 2807 . . . . . . . . . . . . . . . . 17 ((𝜑𝑣𝐴) → 𝐷 = ((𝐺𝑣) + (𝐷 − (𝐺𝑣))))
6867fveq2d 6653 . . . . . . . . . . . . . . . 16 ((𝜑𝑣𝐴) → (abs‘𝐷) = (abs‘((𝐺𝑣) + (𝐷 − (𝐺𝑣)))))
6954, 57subcld 10990 . . . . . . . . . . . . . . . . 17 ((𝜑𝑣𝐴) → (𝐷 − (𝐺𝑣)) ∈ ℂ)
7057, 69abstrid 14812 . . . . . . . . . . . . . . . 16 ((𝜑𝑣𝐴) → (abs‘((𝐺𝑣) + (𝐷 − (𝐺𝑣)))) ≤ ((abs‘(𝐺𝑣)) + (abs‘(𝐷 − (𝐺𝑣)))))
7168, 70eqbrtrd 5055 . . . . . . . . . . . . . . 15 ((𝜑𝑣𝐴) → (abs‘𝐷) ≤ ((abs‘(𝐺𝑣)) + (abs‘(𝐷 − (𝐺𝑣)))))
72713adant3 1129 . . . . . . . . . . . . . 14 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (abs‘𝐷) ≤ ((abs‘(𝐺𝑣)) + (abs‘(𝐷 − (𝐺𝑣)))))
7360, 58abssubd 14809 . . . . . . . . . . . . . . . 16 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (abs‘(𝐷 − (𝐺𝑣))) = (abs‘((𝐺𝑣) − 𝐷)))
74 simp3 1135 . . . . . . . . . . . . . . . 16 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2))
7573, 74eqbrtrd 5055 . . . . . . . . . . . . . . 15 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (abs‘(𝐷 − (𝐺𝑣))) < ((abs‘𝐷) / 2))
7662, 64, 59, 75ltadd2dd 10792 . . . . . . . . . . . . . 14 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → ((abs‘(𝐺𝑣)) + (abs‘(𝐷 − (𝐺𝑣)))) < ((abs‘(𝐺𝑣)) + ((abs‘𝐷) / 2)))
7756, 63, 65, 72, 76lelttrd 10791 . . . . . . . . . . . . 13 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (abs‘𝐷) < ((abs‘(𝐺𝑣)) + ((abs‘𝐷) / 2)))
7857abscld 14792 . . . . . . . . . . . . . . 15 ((𝜑𝑣𝐴) → (abs‘(𝐺𝑣)) ∈ ℝ)
79783adant3 1129 . . . . . . . . . . . . . 14 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (abs‘(𝐺𝑣)) ∈ ℝ)
8056, 64, 79ltsubaddd 11229 . . . . . . . . . . . . 13 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → (((abs‘𝐷) − ((abs‘𝐷) / 2)) < (abs‘(𝐺𝑣)) ↔ (abs‘𝐷) < ((abs‘(𝐺𝑣)) + ((abs‘𝐷) / 2))))
8177, 80mpbird 260 . . . . . . . . . . . 12 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → ((abs‘𝐷) − ((abs‘𝐷) / 2)) < (abs‘(𝐺𝑣)))
8253, 81eqbrtrd 5055 . . . . . . . . . . 11 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → ((abs‘𝐷) − (abs‘(𝐷 / 2))) < (abs‘(𝐺𝑣)))
8350, 82eqbrtrd 5055 . . . . . . . . . 10 ((𝜑𝑣𝐴 ∧ (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))
8426, 27, 30, 83syl3anc 1368 . . . . . . . . 9 ((((𝜑𝑧 ∈ ℝ+) ∧ (𝑣𝐴 → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2))) ∧ 𝑣𝐴) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧)) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))
85843exp1 1349 . . . . . . . 8 ((𝜑𝑧 ∈ ℝ+) → ((𝑣𝐴 → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2))) → (𝑣𝐴 → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))))))
8685ralimdv2 3146 . . . . . . 7 ((𝜑𝑧 ∈ ℝ+) → (∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))))
8786reximdva 3236 . . . . . 6 (𝜑 → (∃𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → (abs‘((𝐺𝑣) − 𝐷)) < ((abs‘𝐷) / 2)) → ∃𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))))
8825, 87mpd 15 . . . . 5 (𝜑 → ∃𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))))
8988adantr 484 . . . 4 ((𝜑𝑦 ∈ ℝ+) → ∃𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))))
90 simpr 488 . . . . . . . . 9 ((𝜑𝑦 ∈ ℝ+) → 𝑦 ∈ ℝ+)
9117adantr 484 . . . . . . . . . . 11 ((𝜑𝑦 ∈ ℝ+) → 𝐷 ∈ ℂ)
9218adantr 484 . . . . . . . . . . 11 ((𝜑𝑦 ∈ ℝ+) → 𝐷 ≠ 0)
9391, 92absrpcld 14804 . . . . . . . . . 10 ((𝜑𝑦 ∈ ℝ+) → (abs‘𝐷) ∈ ℝ+)
9493rphalfcld 12435 . . . . . . . . 9 ((𝜑𝑦 ∈ ℝ+) → ((abs‘𝐷) / 2) ∈ ℝ+)
9590, 94rpmulcld 12439 . . . . . . . 8 ((𝜑𝑦 ∈ ℝ+) → (𝑦 · ((abs‘𝐷) / 2)) ∈ ℝ+)
9695ex 416 . . . . . . . . 9 (𝜑 → (𝑦 ∈ ℝ+ → (𝑦 · ((abs‘𝐷) / 2)) ∈ ℝ+))
9796imdistani 572 . . . . . . . 8 ((𝜑𝑦 ∈ ℝ+) → (𝜑 ∧ (𝑦 · ((abs‘𝐷) / 2)) ∈ ℝ+))
98 eleq1 2880 . . . . . . . . . . 11 (𝑤 = (𝑦 · ((abs‘𝐷) / 2)) → (𝑤 ∈ ℝ+ ↔ (𝑦 · ((abs‘𝐷) / 2)) ∈ ℝ+))
9998anbi2d 631 . . . . . . . . . 10 (𝑤 = (𝑦 · ((abs‘𝐷) / 2)) → ((𝜑𝑤 ∈ ℝ+) ↔ (𝜑 ∧ (𝑦 · ((abs‘𝐷) / 2)) ∈ ℝ+)))
100 breq2 5037 . . . . . . . . . . . 12 (𝑤 = (𝑦 · ((abs‘𝐷) / 2)) → ((abs‘((𝐹𝑣) − 0)) < 𝑤 ↔ (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))))
101100imbi2d 344 . . . . . . . . . . 11 (𝑤 = (𝑦 · ((abs‘𝐷) / 2)) → (((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < 𝑤) ↔ ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))))
102101rexralbidv 3263 . . . . . . . . . 10 (𝑤 = (𝑦 · ((abs‘𝐷) / 2)) → (∃𝑢 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < 𝑤) ↔ ∃𝑢 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))))
10399, 102imbi12d 348 . . . . . . . . 9 (𝑤 = (𝑦 · ((abs‘𝐷) / 2)) → (((𝜑𝑤 ∈ ℝ+) → ∃𝑢 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < 𝑤)) ↔ ((𝜑 ∧ (𝑦 · ((abs‘𝐷) / 2)) ∈ ℝ+) → ∃𝑢 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))))))
1048, 7fmptd 6859 . . . . . . . . . . . . 13 (𝜑𝐹:𝐴⟶ℂ)
105104, 10, 13ellimc3 24486 . . . . . . . . . . . 12 (𝜑 → (0 ∈ (𝐹 lim 𝐸) ↔ (0 ∈ ℂ ∧ ∀𝑤 ∈ ℝ+𝑢 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < 𝑤))))
1069, 105mpbid 235 . . . . . . . . . . 11 (𝜑 → (0 ∈ ℂ ∧ ∀𝑤 ∈ ℝ+𝑢 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < 𝑤)))
107106simprd 499 . . . . . . . . . 10 (𝜑 → ∀𝑤 ∈ ℝ+𝑢 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < 𝑤))
108107r19.21bi 3176 . . . . . . . . 9 ((𝜑𝑤 ∈ ℝ+) → ∃𝑢 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < 𝑤))
109103, 108vtoclg 3518 . . . . . . . 8 ((𝑦 · ((abs‘𝐷) / 2)) ∈ ℝ+ → ((𝜑 ∧ (𝑦 · ((abs‘𝐷) / 2)) ∈ ℝ+) → ∃𝑢 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))))
11095, 97, 109sylc 65 . . . . . . 7 ((𝜑𝑦 ∈ ℝ+) → ∃𝑢 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))))
1111103ad2ant1 1130 . . . . . 6 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) → ∃𝑢 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))))
112 simp12 1201 . . . . . . . . 9 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) → 𝑧 ∈ ℝ+)
113 simp2 1134 . . . . . . . . 9 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) → 𝑢 ∈ ℝ+)
114112, 113ifcld 4473 . . . . . . . 8 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) → if(𝑧𝑢, 𝑧, 𝑢) ∈ ℝ+)
115 nfv 1915 . . . . . . . . . . 11 𝑣(𝜑𝑦 ∈ ℝ+)
116 nfv 1915 . . . . . . . . . . 11 𝑣 𝑧 ∈ ℝ+
117 nfra1 3186 . . . . . . . . . . 11 𝑣𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))
118115, 116, 117nf3an 1902 . . . . . . . . . 10 𝑣((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))))
119 nfv 1915 . . . . . . . . . 10 𝑣 𝑢 ∈ ℝ+
120 nfra1 3186 . . . . . . . . . 10 𝑣𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))
121118, 119, 120nf3an 1902 . . . . . . . . 9 𝑣(((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))))
122 simp111 1299 . . . . . . . . . . . . 13 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → (𝜑𝑦 ∈ ℝ+))
123 simp112 1300 . . . . . . . . . . . . 13 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → 𝑧 ∈ ℝ+)
124 simp12 1201 . . . . . . . . . . . . 13 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → 𝑢 ∈ ℝ+)
125122, 123, 124jca31 518 . . . . . . . . . . . 12 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → (((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+) ∧ 𝑢 ∈ ℝ+))
126 simp2 1134 . . . . . . . . . . . 12 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → 𝑣𝐴)
127 simp3l 1198 . . . . . . . . . . . 12 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → 𝑣𝐸)
128125, 126, 127jca31 518 . . . . . . . . . . 11 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+) ∧ 𝑢 ∈ ℝ+) ∧ 𝑣𝐴) ∧ 𝑣𝐸))
12910adantr 484 . . . . . . . . . . . . . . . . . . 19 ((𝜑𝑦 ∈ ℝ+) → 𝐴 ⊆ ℂ)
1301293ad2ant1 1130 . . . . . . . . . . . . . . . . . 18 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) → 𝐴 ⊆ ℂ)
1311303ad2ant1 1130 . . . . . . . . . . . . . . . . 17 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) → 𝐴 ⊆ ℂ)
1321313ad2ant1 1130 . . . . . . . . . . . . . . . 16 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → 𝐴 ⊆ ℂ)
133132, 126sseldd 3919 . . . . . . . . . . . . . . 15 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → 𝑣 ∈ ℂ)
13413adantr 484 . . . . . . . . . . . . . . . . . 18 ((𝜑𝑦 ∈ ℝ+) → 𝐸 ∈ ℂ)
1351343ad2ant1 1130 . . . . . . . . . . . . . . . . 17 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) → 𝐸 ∈ ℂ)
1361353ad2ant1 1130 . . . . . . . . . . . . . . . 16 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) → 𝐸 ∈ ℂ)
1371363ad2ant1 1130 . . . . . . . . . . . . . . 15 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → 𝐸 ∈ ℂ)
138133, 137subcld 10990 . . . . . . . . . . . . . 14 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → (𝑣𝐸) ∈ ℂ)
139138abscld 14792 . . . . . . . . . . . . 13 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → (abs‘(𝑣𝐸)) ∈ ℝ)
140123rpred 12423 . . . . . . . . . . . . . 14 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → 𝑧 ∈ ℝ)
141124rpred 12423 . . . . . . . . . . . . . 14 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → 𝑢 ∈ ℝ)
142140, 141ifcld 4473 . . . . . . . . . . . . 13 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → if(𝑧𝑢, 𝑧, 𝑢) ∈ ℝ)
143 simp3r 1199 . . . . . . . . . . . . 13 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))
144 min1 12574 . . . . . . . . . . . . . 14 ((𝑧 ∈ ℝ ∧ 𝑢 ∈ ℝ) → if(𝑧𝑢, 𝑧, 𝑢) ≤ 𝑧)
145140, 141, 144syl2anc 587 . . . . . . . . . . . . 13 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → if(𝑧𝑢, 𝑧, 𝑢) ≤ 𝑧)
146139, 142, 140, 143, 145ltletrd 10793 . . . . . . . . . . . 12 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → (abs‘(𝑣𝐸)) < 𝑧)
147 simp113 1301 . . . . . . . . . . . . 13 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))))
148 rspa 3174 . . . . . . . . . . . . 13 ((∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ 𝑣𝐴) → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))))
149147, 126, 148syl2anc 587 . . . . . . . . . . . 12 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))))
150127, 146, 149mp2and 698 . . . . . . . . . . 11 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))
151 simp13 1202 . . . . . . . . . . . . 13 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))))
152 rspa 3174 . . . . . . . . . . . . 13 ((∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))) ∧ 𝑣𝐴) → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))))
153151, 126, 152syl2anc 587 . . . . . . . . . . . 12 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))))
154 min2 12575 . . . . . . . . . . . . . . 15 ((𝑧 ∈ ℝ ∧ 𝑢 ∈ ℝ) → if(𝑧𝑢, 𝑧, 𝑢) ≤ 𝑢)
155140, 141, 154syl2anc 587 . . . . . . . . . . . . . 14 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → if(𝑧𝑢, 𝑧, 𝑢) ≤ 𝑢)
156139, 142, 141, 143, 155ltletrd 10793 . . . . . . . . . . . . 13 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → (abs‘(𝑣𝐸)) < 𝑢)
157127, 156jca 515 . . . . . . . . . . . 12 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢))
158122simpld 498 . . . . . . . . . . . . . . 15 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → 𝜑)
1591583ad2ant1 1130 . . . . . . . . . . . . . 14 ((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) ∧ ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢)) → 𝜑)
160 simp12 1201 . . . . . . . . . . . . . 14 ((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) ∧ ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢)) → 𝑣𝐴)
161 nfv 1915 . . . . . . . . . . . . . . . . . . 19 𝑥(𝜑𝑣𝐴)
162 nfmpt1 5131 . . . . . . . . . . . . . . . . . . . . . 22 𝑥(𝑥𝐴𝐵)
1637, 162nfcxfr 2956 . . . . . . . . . . . . . . . . . . . . 21 𝑥𝐹
164 nfcv 2958 . . . . . . . . . . . . . . . . . . . . 21 𝑥𝑣
165163, 164nffv 6659 . . . . . . . . . . . . . . . . . . . 20 𝑥(𝐹𝑣)
166165nfel1 2974 . . . . . . . . . . . . . . . . . . 19 𝑥(𝐹𝑣) ∈ ℂ
167161, 166nfim 1897 . . . . . . . . . . . . . . . . . 18 𝑥((𝜑𝑣𝐴) → (𝐹𝑣) ∈ ℂ)
168 eleq1 2880 . . . . . . . . . . . . . . . . . . . 20 (𝑥 = 𝑣 → (𝑥𝐴𝑣𝐴))
169168anbi2d 631 . . . . . . . . . . . . . . . . . . 19 (𝑥 = 𝑣 → ((𝜑𝑥𝐴) ↔ (𝜑𝑣𝐴)))
170 fveq2 6649 . . . . . . . . . . . . . . . . . . . 20 (𝑥 = 𝑣 → (𝐹𝑥) = (𝐹𝑣))
171170eleq1d 2877 . . . . . . . . . . . . . . . . . . 19 (𝑥 = 𝑣 → ((𝐹𝑥) ∈ ℂ ↔ (𝐹𝑣) ∈ ℂ))
172169, 171imbi12d 348 . . . . . . . . . . . . . . . . . 18 (𝑥 = 𝑣 → (((𝜑𝑥𝐴) → (𝐹𝑥) ∈ ℂ) ↔ ((𝜑𝑣𝐴) → (𝐹𝑣) ∈ ℂ)))
173 simpr 488 . . . . . . . . . . . . . . . . . . . 20 ((𝜑𝑥𝐴) → 𝑥𝐴)
1747fvmpt2 6760 . . . . . . . . . . . . . . . . . . . 20 ((𝑥𝐴𝐵 ∈ ℂ) → (𝐹𝑥) = 𝐵)
175173, 8, 174syl2anc 587 . . . . . . . . . . . . . . . . . . 19 ((𝜑𝑥𝐴) → (𝐹𝑥) = 𝐵)
176175, 8eqeltrd 2893 . . . . . . . . . . . . . . . . . 18 ((𝜑𝑥𝐴) → (𝐹𝑥) ∈ ℂ)
177167, 172, 176chvarfv 2241 . . . . . . . . . . . . . . . . 17 ((𝜑𝑣𝐴) → (𝐹𝑣) ∈ ℂ)
178177subid1d 10979 . . . . . . . . . . . . . . . 16 ((𝜑𝑣𝐴) → ((𝐹𝑣) − 0) = (𝐹𝑣))
179178eqcomd 2807 . . . . . . . . . . . . . . 15 ((𝜑𝑣𝐴) → (𝐹𝑣) = ((𝐹𝑣) − 0))
180179fveq2d 6653 . . . . . . . . . . . . . 14 ((𝜑𝑣𝐴) → (abs‘(𝐹𝑣)) = (abs‘((𝐹𝑣) − 0)))
181159, 160, 180syl2anc 587 . . . . . . . . . . . . 13 ((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) ∧ ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢)) → (abs‘(𝐹𝑣)) = (abs‘((𝐹𝑣) − 0)))
182 simp3 1135 . . . . . . . . . . . . . 14 ((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) ∧ ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢)) → (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢))
183 simp2 1134 . . . . . . . . . . . . . 14 ((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) ∧ ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢)) → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))))
184182, 183mpd 15 . . . . . . . . . . . . 13 ((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) ∧ ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢)) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))
185181, 184eqbrtrd 5055 . . . . . . . . . . . 12 ((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) ∧ ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))) ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢)) → (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2)))
186153, 157, 185mpd3an23 1460 . . . . . . . . . . 11 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2)))
187 simp-7l 788 . . . . . . . . . . . . 13 ((((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+) ∧ 𝑢 ∈ ℝ+) ∧ 𝑣𝐴) ∧ 𝑣𝐸) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → 𝜑)
188 simp-4r 783 . . . . . . . . . . . . 13 ((((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+) ∧ 𝑢 ∈ ℝ+) ∧ 𝑣𝐴) ∧ 𝑣𝐸) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → 𝑣𝐴)
189 eldifsni 4686 . . . . . . . . . . . . . . . . . . . 20 (𝐶 ∈ (ℂ ∖ {0}) → 𝐶 ≠ 0)
1903, 189syl 17 . . . . . . . . . . . . . . . . . . 19 ((𝜑𝑥𝐴) → 𝐶 ≠ 0)
1918, 4, 190divcld 11409 . . . . . . . . . . . . . . . . . 18 ((𝜑𝑥𝐴) → (𝐵 / 𝐶) ∈ ℂ)
192 0ellimcdiv.h . . . . . . . . . . . . . . . . . 18 𝐻 = (𝑥𝐴 ↦ (𝐵 / 𝐶))
193191, 192fmptd 6859 . . . . . . . . . . . . . . . . 17 (𝜑𝐻:𝐴⟶ℂ)
194193ffvelrnda 6832 . . . . . . . . . . . . . . . 16 ((𝜑𝑣𝐴) → (𝐻𝑣) ∈ ℂ)
195194subid1d 10979 . . . . . . . . . . . . . . 15 ((𝜑𝑣𝐴) → ((𝐻𝑣) − 0) = (𝐻𝑣))
196 nfmpt1 5131 . . . . . . . . . . . . . . . . . . . 20 𝑥(𝑥𝐴 ↦ (𝐵 / 𝐶))
197192, 196nfcxfr 2956 . . . . . . . . . . . . . . . . . . 19 𝑥𝐻
198197, 164nffv 6659 . . . . . . . . . . . . . . . . . 18 𝑥(𝐻𝑣)
199 nfcv 2958 . . . . . . . . . . . . . . . . . . 19 𝑥 /
200 nfmpt1 5131 . . . . . . . . . . . . . . . . . . . . 21 𝑥(𝑥𝐴𝐶)
2015, 200nfcxfr 2956 . . . . . . . . . . . . . . . . . . . 20 𝑥𝐺
202201, 164nffv 6659 . . . . . . . . . . . . . . . . . . 19 𝑥(𝐺𝑣)
203165, 199, 202nfov 7169 . . . . . . . . . . . . . . . . . 18 𝑥((𝐹𝑣) / (𝐺𝑣))
204198, 203nfeq 2971 . . . . . . . . . . . . . . . . 17 𝑥(𝐻𝑣) = ((𝐹𝑣) / (𝐺𝑣))
205161, 204nfim 1897 . . . . . . . . . . . . . . . 16 𝑥((𝜑𝑣𝐴) → (𝐻𝑣) = ((𝐹𝑣) / (𝐺𝑣)))
206 fveq2 6649 . . . . . . . . . . . . . . . . . 18 (𝑥 = 𝑣 → (𝐻𝑥) = (𝐻𝑣))
207 fveq2 6649 . . . . . . . . . . . . . . . . . . 19 (𝑥 = 𝑣 → (𝐺𝑥) = (𝐺𝑣))
208170, 207oveq12d 7157 . . . . . . . . . . . . . . . . . 18 (𝑥 = 𝑣 → ((𝐹𝑥) / (𝐺𝑥)) = ((𝐹𝑣) / (𝐺𝑣)))
209206, 208eqeq12d 2817 . . . . . . . . . . . . . . . . 17 (𝑥 = 𝑣 → ((𝐻𝑥) = ((𝐹𝑥) / (𝐺𝑥)) ↔ (𝐻𝑣) = ((𝐹𝑣) / (𝐺𝑣))))
210169, 209imbi12d 348 . . . . . . . . . . . . . . . 16 (𝑥 = 𝑣 → (((𝜑𝑥𝐴) → (𝐻𝑥) = ((𝐹𝑥) / (𝐺𝑥))) ↔ ((𝜑𝑣𝐴) → (𝐻𝑣) = ((𝐹𝑣) / (𝐺𝑣)))))
211192fvmpt2 6760 . . . . . . . . . . . . . . . . . 18 ((𝑥𝐴 ∧ (𝐵 / 𝐶) ∈ ℂ) → (𝐻𝑥) = (𝐵 / 𝐶))
212173, 191, 211syl2anc 587 . . . . . . . . . . . . . . . . 17 ((𝜑𝑥𝐴) → (𝐻𝑥) = (𝐵 / 𝐶))
213175eqcomd 2807 . . . . . . . . . . . . . . . . . 18 ((𝜑𝑥𝐴) → 𝐵 = (𝐹𝑥))
2145fvmpt2 6760 . . . . . . . . . . . . . . . . . . . 20 ((𝑥𝐴𝐶 ∈ (ℂ ∖ {0})) → (𝐺𝑥) = 𝐶)
215173, 3, 214syl2anc 587 . . . . . . . . . . . . . . . . . . 19 ((𝜑𝑥𝐴) → (𝐺𝑥) = 𝐶)
216215eqcomd 2807 . . . . . . . . . . . . . . . . . 18 ((𝜑𝑥𝐴) → 𝐶 = (𝐺𝑥))
217213, 216oveq12d 7157 . . . . . . . . . . . . . . . . 17 ((𝜑𝑥𝐴) → (𝐵 / 𝐶) = ((𝐹𝑥) / (𝐺𝑥)))
218212, 217eqtrd 2836 . . . . . . . . . . . . . . . 16 ((𝜑𝑥𝐴) → (𝐻𝑥) = ((𝐹𝑥) / (𝐺𝑥)))
219205, 210, 218chvarfv 2241 . . . . . . . . . . . . . . 15 ((𝜑𝑣𝐴) → (𝐻𝑣) = ((𝐹𝑣) / (𝐺𝑣)))
220195, 219eqtrd 2836 . . . . . . . . . . . . . 14 ((𝜑𝑣𝐴) → ((𝐻𝑣) − 0) = ((𝐹𝑣) / (𝐺𝑣)))
221220fveq2d 6653 . . . . . . . . . . . . 13 ((𝜑𝑣𝐴) → (abs‘((𝐻𝑣) − 0)) = (abs‘((𝐹𝑣) / (𝐺𝑣))))
222187, 188, 221syl2anc 587 . . . . . . . . . . . 12 ((((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+) ∧ 𝑢 ∈ ℝ+) ∧ 𝑣𝐴) ∧ 𝑣𝐸) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → (abs‘((𝐻𝑣) − 0)) = (abs‘((𝐹𝑣) / (𝐺𝑣))))
223 simp-6l 786 . . . . . . . . . . . . . 14 ((((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+) ∧ 𝑢 ∈ ℝ+) ∧ 𝑣𝐴) ∧ 𝑣𝐸) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → (𝜑𝑦 ∈ ℝ+))
224223, 188jca 515 . . . . . . . . . . . . 13 ((((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+) ∧ 𝑢 ∈ ℝ+) ∧ 𝑣𝐴) ∧ 𝑣𝐸) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → ((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴))
225 simplr 768 . . . . . . . . . . . . 13 ((((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+) ∧ 𝑢 ∈ ℝ+) ∧ 𝑣𝐴) ∧ 𝑣𝐸) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))
226 simpr 488 . . . . . . . . . . . . 13 ((((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+) ∧ 𝑢 ∈ ℝ+) ∧ 𝑣𝐴) ∧ 𝑣𝐸) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2)))
227 nfcv 2958 . . . . . . . . . . . . . . . . . . . 20 𝑥0
228202, 227nfne 3090 . . . . . . . . . . . . . . . . . . 19 𝑥(𝐺𝑣) ≠ 0
229161, 228nfim 1897 . . . . . . . . . . . . . . . . . 18 𝑥((𝜑𝑣𝐴) → (𝐺𝑣) ≠ 0)
230207neeq1d 3049 . . . . . . . . . . . . . . . . . . 19 (𝑥 = 𝑣 → ((𝐺𝑥) ≠ 0 ↔ (𝐺𝑣) ≠ 0))
231169, 230imbi12d 348 . . . . . . . . . . . . . . . . . 18 (𝑥 = 𝑣 → (((𝜑𝑥𝐴) → (𝐺𝑥) ≠ 0) ↔ ((𝜑𝑣𝐴) → (𝐺𝑣) ≠ 0)))
232215, 190eqnetrd 3057 . . . . . . . . . . . . . . . . . 18 ((𝜑𝑥𝐴) → (𝐺𝑥) ≠ 0)
233229, 231, 232chvarfv 2241 . . . . . . . . . . . . . . . . 17 ((𝜑𝑣𝐴) → (𝐺𝑣) ≠ 0)
234177, 57, 233absdivd 14811 . . . . . . . . . . . . . . . 16 ((𝜑𝑣𝐴) → (abs‘((𝐹𝑣) / (𝐺𝑣))) = ((abs‘(𝐹𝑣)) / (abs‘(𝐺𝑣))))
235234adantlr 714 . . . . . . . . . . . . . . 15 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → (abs‘((𝐹𝑣) / (𝐺𝑣))) = ((abs‘(𝐹𝑣)) / (abs‘(𝐺𝑣))))
236235ad2antrr 725 . . . . . . . . . . . . . 14 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → (abs‘((𝐹𝑣) / (𝐺𝑣))) = ((abs‘(𝐹𝑣)) / (abs‘(𝐺𝑣))))
237177abscld 14792 . . . . . . . . . . . . . . . . . 18 ((𝜑𝑣𝐴) → (abs‘(𝐹𝑣)) ∈ ℝ)
23857, 233absne0d 14803 . . . . . . . . . . . . . . . . . 18 ((𝜑𝑣𝐴) → (abs‘(𝐺𝑣)) ≠ 0)
239237, 78, 238redivcld 11461 . . . . . . . . . . . . . . . . 17 ((𝜑𝑣𝐴) → ((abs‘(𝐹𝑣)) / (abs‘(𝐺𝑣))) ∈ ℝ)
240239adantlr 714 . . . . . . . . . . . . . . . 16 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → ((abs‘(𝐹𝑣)) / (abs‘(𝐺𝑣))) ∈ ℝ)
241240ad2antrr 725 . . . . . . . . . . . . . . 15 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → ((abs‘(𝐹𝑣)) / (abs‘(𝐺𝑣))) ∈ ℝ)
242 rpre 12389 . . . . . . . . . . . . . . . . . . 19 (𝑦 ∈ ℝ+𝑦 ∈ ℝ)
243242ad2antlr 726 . . . . . . . . . . . . . . . . . 18 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → 𝑦 ∈ ℝ)
24420rpred 12423 . . . . . . . . . . . . . . . . . . 19 (𝜑 → ((abs‘𝐷) / 2) ∈ ℝ)
245244ad2antrr 725 . . . . . . . . . . . . . . . . . 18 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → ((abs‘𝐷) / 2) ∈ ℝ)
246243, 245remulcld 10664 . . . . . . . . . . . . . . . . 17 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → (𝑦 · ((abs‘𝐷) / 2)) ∈ ℝ)
247246ad2antrr 725 . . . . . . . . . . . . . . . 16 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → (𝑦 · ((abs‘𝐷) / 2)) ∈ ℝ)
24857, 233absrpcld 14804 . . . . . . . . . . . . . . . . . 18 ((𝜑𝑣𝐴) → (abs‘(𝐺𝑣)) ∈ ℝ+)
249248adantlr 714 . . . . . . . . . . . . . . . . 17 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → (abs‘(𝐺𝑣)) ∈ ℝ+)
250249ad2antrr 725 . . . . . . . . . . . . . . . 16 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → (abs‘(𝐺𝑣)) ∈ ℝ+)
251247, 250rerpdivcld 12454 . . . . . . . . . . . . . . 15 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → ((𝑦 · ((abs‘𝐷) / 2)) / (abs‘(𝐺𝑣))) ∈ ℝ)
252243ad2antrr 725 . . . . . . . . . . . . . . 15 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → 𝑦 ∈ ℝ)
253 simp-4l 782 . . . . . . . . . . . . . . . . 17 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → 𝜑)
254 simpllr 775 . . . . . . . . . . . . . . . . 17 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → 𝑣𝐴)
255253, 254, 237syl2anc 587 . . . . . . . . . . . . . . . 16 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → (abs‘(𝐹𝑣)) ∈ ℝ)
256 simpr 488 . . . . . . . . . . . . . . . 16 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2)))
257255, 247, 250, 256ltdiv1dd 12480 . . . . . . . . . . . . . . 15 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → ((abs‘(𝐹𝑣)) / (abs‘(𝐺𝑣))) < ((𝑦 · ((abs‘𝐷) / 2)) / (abs‘(𝐺𝑣))))
258243recnd 10662 . . . . . . . . . . . . . . . . . . . 20 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → 𝑦 ∈ ℂ)
25947ad2antrr 725 . . . . . . . . . . . . . . . . . . . 20 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → ((abs‘𝐷) / 2) ∈ ℂ)
260249rpcnd 12425 . . . . . . . . . . . . . . . . . . . 20 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → (abs‘(𝐺𝑣)) ∈ ℂ)
261238adantlr 714 . . . . . . . . . . . . . . . . . . . 20 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → (abs‘(𝐺𝑣)) ≠ 0)
262258, 259, 260, 261divassd 11444 . . . . . . . . . . . . . . . . . . 19 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → ((𝑦 · ((abs‘𝐷) / 2)) / (abs‘(𝐺𝑣))) = (𝑦 · (((abs‘𝐷) / 2) / (abs‘(𝐺𝑣)))))
263262adantr 484 . . . . . . . . . . . . . . . . . 18 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → ((𝑦 · ((abs‘𝐷) / 2)) / (abs‘(𝐺𝑣))) = (𝑦 · (((abs‘𝐷) / 2) / (abs‘(𝐺𝑣)))))
264245, 249rerpdivcld 12454 . . . . . . . . . . . . . . . . . . . 20 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → (((abs‘𝐷) / 2) / (abs‘(𝐺𝑣))) ∈ ℝ)
265264adantr 484 . . . . . . . . . . . . . . . . . . 19 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → (((abs‘𝐷) / 2) / (abs‘(𝐺𝑣))) ∈ ℝ)
266 1red 10635 . . . . . . . . . . . . . . . . . . 19 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → 1 ∈ ℝ)
267 simpllr 775 . . . . . . . . . . . . . . . . . . 19 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → 𝑦 ∈ ℝ+)
268244ad2antrr 725 . . . . . . . . . . . . . . . . . . . . 21 (((𝜑𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → ((abs‘𝐷) / 2) ∈ ℝ)
269 1rp 12385 . . . . . . . . . . . . . . . . . . . . . 22 1 ∈ ℝ+
270269a1i 11 . . . . . . . . . . . . . . . . . . . . 21 (((𝜑𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → 1 ∈ ℝ+)
271248adantr 484 . . . . . . . . . . . . . . . . . . . . 21 (((𝜑𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → (abs‘(𝐺𝑣)) ∈ ℝ+)
27247div1d 11401 . . . . . . . . . . . . . . . . . . . . . . 23 (𝜑 → (((abs‘𝐷) / 2) / 1) = ((abs‘𝐷) / 2))
273272ad2antrr 725 . . . . . . . . . . . . . . . . . . . . . 22 (((𝜑𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → (((abs‘𝐷) / 2) / 1) = ((abs‘𝐷) / 2))
274 simpr 488 . . . . . . . . . . . . . . . . . . . . . 22 (((𝜑𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))
275273, 274eqbrtrd 5055 . . . . . . . . . . . . . . . . . . . . 21 (((𝜑𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → (((abs‘𝐷) / 2) / 1) < (abs‘(𝐺𝑣)))
276268, 270, 271, 275ltdiv23d 12490 . . . . . . . . . . . . . . . . . . . 20 (((𝜑𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → (((abs‘𝐷) / 2) / (abs‘(𝐺𝑣))) < 1)
277276adantllr 718 . . . . . . . . . . . . . . . . . . 19 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → (((abs‘𝐷) / 2) / (abs‘(𝐺𝑣))) < 1)
278265, 266, 267, 277ltmul2dd 12479 . . . . . . . . . . . . . . . . . 18 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → (𝑦 · (((abs‘𝐷) / 2) / (abs‘(𝐺𝑣)))) < (𝑦 · 1))
279263, 278eqbrtrd 5055 . . . . . . . . . . . . . . . . 17 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → ((𝑦 · ((abs‘𝐷) / 2)) / (abs‘(𝐺𝑣))) < (𝑦 · 1))
280258mulid1d 10651 . . . . . . . . . . . . . . . . . 18 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) → (𝑦 · 1) = 𝑦)
281280adantr 484 . . . . . . . . . . . . . . . . 17 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → (𝑦 · 1) = 𝑦)
282279, 281breqtrd 5059 . . . . . . . . . . . . . . . 16 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → ((𝑦 · ((abs‘𝐷) / 2)) / (abs‘(𝐺𝑣))) < 𝑦)
283282adantr 484 . . . . . . . . . . . . . . 15 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → ((𝑦 · ((abs‘𝐷) / 2)) / (abs‘(𝐺𝑣))) < 𝑦)
284241, 251, 252, 257, 283lttrd 10794 . . . . . . . . . . . . . 14 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → ((abs‘(𝐹𝑣)) / (abs‘(𝐺𝑣))) < 𝑦)
285236, 284eqbrtrd 5055 . . . . . . . . . . . . 13 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑣𝐴) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → (abs‘((𝐹𝑣) / (𝐺𝑣))) < 𝑦)
286224, 225, 226, 285syl21anc 836 . . . . . . . . . . . 12 ((((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+) ∧ 𝑢 ∈ ℝ+) ∧ 𝑣𝐴) ∧ 𝑣𝐸) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → (abs‘((𝐹𝑣) / (𝐺𝑣))) < 𝑦)
287222, 286eqbrtrd 5055 . . . . . . . . . . 11 ((((((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+) ∧ 𝑢 ∈ ℝ+) ∧ 𝑣𝐴) ∧ 𝑣𝐸) ∧ ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) ∧ (abs‘(𝐹𝑣)) < (𝑦 · ((abs‘𝐷) / 2))) → (abs‘((𝐻𝑣) − 0)) < 𝑦)
288128, 150, 186, 287syl21anc 836 . . . . . . . . . 10 (((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) ∧ 𝑣𝐴 ∧ (𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢))) → (abs‘((𝐻𝑣) − 0)) < 𝑦)
2892883exp 1116 . . . . . . . . 9 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) → (𝑣𝐴 → ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢)) → (abs‘((𝐻𝑣) − 0)) < 𝑦)))
290121, 289ralrimi 3183 . . . . . . . 8 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) → ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢)) → (abs‘((𝐻𝑣) − 0)) < 𝑦))
291 brimralrspcev 5094 . . . . . . . 8 ((if(𝑧𝑢, 𝑧, 𝑢) ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < if(𝑧𝑢, 𝑧, 𝑢)) → (abs‘((𝐻𝑣) − 0)) < 𝑦)) → ∃𝑤 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑤) → (abs‘((𝐻𝑣) − 0)) < 𝑦))
292114, 290, 291syl2anc 587 . . . . . . 7 ((((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) ∧ 𝑢 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2)))) → ∃𝑤 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑤) → (abs‘((𝐻𝑣) − 0)) < 𝑦))
293292rexlimdv3a 3248 . . . . . 6 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) → (∃𝑢 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑢) → (abs‘((𝐹𝑣) − 0)) < (𝑦 · ((abs‘𝐷) / 2))) → ∃𝑤 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑤) → (abs‘((𝐻𝑣) − 0)) < 𝑦)))
294111, 293mpd 15 . . . . 5 (((𝜑𝑦 ∈ ℝ+) ∧ 𝑧 ∈ ℝ+ ∧ ∀𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣)))) → ∃𝑤 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑤) → (abs‘((𝐻𝑣) − 0)) < 𝑦))
295294rexlimdv3a 3248 . . . 4 ((𝜑𝑦 ∈ ℝ+) → (∃𝑧 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑧) → ((abs‘𝐷) / 2) < (abs‘(𝐺𝑣))) → ∃𝑤 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑤) → (abs‘((𝐻𝑣) − 0)) < 𝑦)))
29689, 295mpd 15 . . 3 ((𝜑𝑦 ∈ ℝ+) → ∃𝑤 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑤) → (abs‘((𝐻𝑣) − 0)) < 𝑦))
297296ralrimiva 3152 . 2 (𝜑 → ∀𝑦 ∈ ℝ+𝑤 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑤) → (abs‘((𝐻𝑣) − 0)) < 𝑦))
298193, 10, 13ellimc3 24486 . 2 (𝜑 → (0 ∈ (𝐻 lim 𝐸) ↔ (0 ∈ ℂ ∧ ∀𝑦 ∈ ℝ+𝑤 ∈ ℝ+𝑣𝐴 ((𝑣𝐸 ∧ (abs‘(𝑣𝐸)) < 𝑤) → (abs‘((𝐻𝑣) − 0)) < 𝑦))))
2991, 297, 298mpbir2and 712 1 (𝜑 → 0 ∈ (𝐻 lim 𝐸))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 399 ∧ w3a 1084 = wceq 1538 ∈ wcel 2112 ≠ wne 2990 ∀wral 3109 ∃wrex 3110 ∖ cdif 3881 ⊆ wss 3884 ifcif 4428 {csn 4528 class class class wbr 5033 ↦ cmpt 5113 dom cdm 5523 ⟶wf 6324 ‘cfv 6328 (class class class)co 7139 ℂcc 10528 ℝcr 10529 0cc0 10530 1c1 10531 + caddc 10533 · cmul 10535 < clt 10668 ≤ cle 10669 − cmin 10863 / cdiv 11290 2c2 11684 ℝ+crp 12381 abscabs 14589 limℂ climc 24469 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1797 ax-4 1811 ax-5 1911 ax-6 1970 ax-7 2015 ax-8 2114 ax-9 2122 ax-10 2143 ax-11 2159 ax-12 2176 ax-ext 2773 ax-rep 5157 ax-sep 5170 ax-nul 5177 ax-pow 5234 ax-pr 5298 ax-un 7445 ax-cnex 10586 ax-resscn 10587 ax-1cn 10588 ax-icn 10589 ax-addcl 10590 ax-addrcl 10591 ax-mulcl 10592 ax-mulrcl 10593 ax-mulcom 10594 ax-addass 10595 ax-mulass 10596 ax-distr 10597 ax-i2m1 10598 ax-1ne0 10599 ax-1rid 10600 ax-rnegex 10601 ax-rrecex 10602 ax-cnre 10603 ax-pre-lttri 10604 ax-pre-lttrn 10605 ax-pre-ltadd 10606 ax-pre-mulgt0 10607 ax-pre-sup 10608 This theorem depends on definitions: df-bi 210 df-an 400 df-or 845 df-3or 1085 df-3an 1086 df-tru 1541 df-ex 1782 df-nf 1786 df-sb 2070 df-mo 2601 df-eu 2632 df-clab 2780 df-cleq 2794 df-clel 2873 df-nfc 2941 df-ne 2991 df-nel 3095 df-ral 3114 df-rex 3115 df-reu 3116 df-rmo 3117 df-rab 3118 df-v 3446 df-sbc 3724 df-csb 3832 df-dif 3887 df-un 3889 df-in 3891 df-ss 3901 df-pss 3903 df-nul 4247 df-if 4429 df-pw 4502 df-sn 4529 df-pr 4531 df-tp 4533 df-op 4535 df-uni 4804 df-int 4842 df-iun 4886 df-br 5034 df-opab 5096 df-mpt 5114 df-tr 5140 df-id 5428 df-eprel 5433 df-po 5442 df-so 5443 df-fr 5482 df-we 5484 df-xp 5529 df-rel 5530 df-cnv 5531 df-co 5532 df-dm 5533 df-rn 5534 df-res 5535 df-ima 5536 df-pred 6120 df-ord 6166 df-on 6167 df-lim 6168 df-suc 6169 df-iota 6287 df-fun 6330 df-fn 6331 df-f 6332 df-f1 6333 df-fo 6334 df-f1o 6335 df-fv 6336 df-riota 7097 df-ov 7142 df-oprab 7143 df-mpo 7144 df-om 7565 df-1st 7675 df-2nd 7676 df-wrecs 7934 df-recs 7995 df-rdg 8033 df-1o 8089 df-oadd 8093 df-er 8276 df-map 8395 df-pm 8396 df-en 8497 df-dom 8498 df-sdom 8499 df-fin 8500 df-fi 8863 df-sup 8894 df-inf 8895 df-pnf 10670 df-mnf 10671 df-xr 10672 df-ltxr 10673 df-le 10674 df-sub 10865 df-neg 10866 df-div 11291 df-nn 11630 df-2 11692 df-3 11693 df-4 11694 df-5 11695 df-6 11696 df-7 11697 df-8 11698 df-9 11699 df-n0 11890 df-z 11974 df-dec 12091 df-uz 12236 df-q 12341 df-rp 12382 df-xneg 12499 df-xadd 12500 df-xmul 12501 df-fz 12890 df-seq 13369 df-exp 13430 df-cj 14454 df-re 14455 df-im 14456 df-sqrt 14590 df-abs 14591 df-struct 16481 df-ndx 16482 df-slot 16483 df-base 16485 df-plusg 16574 df-mulr 16575 df-starv 16576 df-tset 16580 df-ple 16581 df-ds 16583 df-unif 16584 df-rest 16692 df-topn 16693 df-topgen 16713 df-psmet 20087 df-xmet 20088 df-met 20089 df-bl 20090 df-mopn 20091 df-cnfld 20096 df-top 21503 df-topon 21520 df-topsp 21542 df-bases 21555 df-cnp 21837 df-xms 22931 df-ms 22932 df-limc 24473 This theorem is referenced by: reclimc 42288
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# Supp#2PROBABILITY_THEORY_FOR_EE_150 - PROBABILITY THEORY...
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1 PROBABILITY THEORY FOR EE 150 Background for coin toss problems and chemical reaction simulations Probability theory is the branch of mathematics concerned with analysis of random phenomena. The central objects of probability theory are random variables, stochastic processes, and events: mathematical abstractions of non-deterministic events or measured quantities that may either be single occurrences or evolve over time in an apparently random fashion. Although an individual coin toss or the roll of a die is a random event, if repeated many times the sequence of random events will exhibit certain statistical patterns, which can be studied and predicted. Two representative mathematical results describing such patterns are the law of large numbers and the central limit theorem. As a mathematical foundation for statistics, probability theory is essential to many human activities that involve quantitative analysis of large sets of data. Methods of probability theory also apply to description of complex systems given only partial knowledge of their state, as in statistical mechanics. A great discovery of twentieth century physics was the probabilistic nature of physical phenomena at atomic scales, described in quantum mechanics. In probability theory, an event is a set of outcomes (a subset of the of a larger set which is usually defined as the set of possible outcomes of interest ) to which a probability is assigned. Typically, when the set of possible outcomes is finite, any subset of the set of possible outcomes is an event ( i . e . all elements of the of the set of possible outcomes are defined as events). Probability plays a role in many sectors of modern life. Most of us play games of chance in some way or other. This may vary from a simple board game to playing games in a casino for money. In corporate life, actuaries are employed to assess risk of almost every kind so that insurance companies make profits on the premiums they set. There are a number of ways to try to determine the probability of an event. Here we use probability trees. This method is useful when we have basic activities with known theoretical probabilities, such as rolling a dice, tossing a coin, drawing a card from a deck or taking a ball from a barrel. When these activities are repeated we can find the probability of an event such as getting three sixes in three rolls of a dice, by constructing a probability tree. Probability trees can also be used in situations where two different basic events are combined and to compute conditional probability. But first we will just be concerned with learning how to construct and use probability trees when one activity is repeated. Along the way one should always look for patterns in your work. Here we will look for generalizations. What will happen when a coin is tossed n times? What is the likelihood of getting precisely one Tail? Two? These generalizations are basically like finding algebraic patterns but involve slightly more sophistication.
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Supp#2PROBABILITY_THEORY_FOR_EE_150 - PROBABILITY THEORY...
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# Review1-4up - 1 STA 2023 c B.Presnell& D.Wackerly Review...
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Unformatted text preview: 1 STA 2023 c B.Presnell & D.Wackerly - Review for Exam I STA 2023 c B.Presnell & D.Wackerly - Review for Exam I 2 Chapter 1 : Chapter 2 : ยก Descriptive (Communicate) and Inferential (estimates, decisions) Statistics (p. 2, 3) ยก ยก Types of Data (p. 8): Quantitative, Qualitative Population (p. 4) ยก ยก Graphical Methods:Stem/Leaf, Histograms, Variable (p. 4) ยก Sample (p. 5) ยก Five Elements of a Statistical Problem (p. 8) Box-plots, outliers, scatter diagrams (p. 28โ30, 68-71, 78-79, ) ยก Measure of Central Tendency (p. 40) โ Where is the โmiddleโ? (p. 41) โ Population Mean : 3. Sample of population units. ยค โ Sample Mean : 2. Speci๏ฌcation of Variables to be investigated ยขยฃ 1. Clear speci๏ฌcation population of interest , โmuโ (p. 42) โ The median (p. 42-43) 4. Inference about population based on info in โ Skewness and Symmetry (p. 45) sample. โ The mode (p. 44) 5. Measure of the goodness or reliability of the inference. 3 STA 2023 c B.Presnell & D.Wackerly - Review for Exam I STA 2023 c B.Presnell & D.Wackerly - Review for Exam I Interpreting the mean and standard deviation ) 10ยก โ The range (p. 51) (p. 65) Empirical Rule (bell-shaped distns) โapprox.โ (p. 57) Miscellaneous ยก Percentiles (p. 64) Quartiles (p. 69) ยก Rare Events and Inference - requires the assess- ยฆ ยฆ # \$! ยง " ยจ ยฆ ยจ ยฉยง (p. 53) (p. 53) ยฅ ยฅ ( ยง % % #! 'ยง & ยจ % ยฅ ยฅ โ Variance or standard deviation LARGE MORE variability. standard deviation ยก Std. Dev. : mean leastโ(p. 56) (p. 52) (p. 52) โ POPULATION Variance : value Tchebysheffโs Theorem (always works) โat ยก Std. Dev. : ยข ยง ยฃ ยฃ Variance : ยก โ SAMPLE score : How about Variability, Spread or Dispersion? ment of how likely or probable an occurrence is. 4 ยก 5 STA 2023 c B.Presnell & D.Wackerly - Review for Exam I STA 2023 c B.Presnell & D.Wackerly - Review for Exam I ยก ยก ยก ยก Probability (p. 102) ยก Sample Point (p. 101) ยก Sample Space, ยก Properties of Probabilities of Simple Events (p. 104) ยก Events (p. 105) ยจ ยก ยฃ Experiment (p. 100) ยก How to ๏ฌnd the probability of an event (p. 106) ยก Intersection and Union of two events (p. 104) ยข ยฃ ยก ยฃ ยจ ยข ยงยก ยฃ ยฆ ยก Additive Rule (p. 117): ยก Conditional probability (p. 122) ยข ยฉยก ยฃ ยก Multiplicative Law (p. 128): # ยก ยฃ ยฉยก ยข ยจ ยข ยข ยก ยจ ยข ยงยก ยฉยฃ ยจยฃ ยฃ ยฃ IW ยก RB DB IB ยข STA 2023 c B.Presnell & D.Wackerly - Review for Exam I 7 STA 2023 c B.Presnell & D.Wackerly - Review for Exam I Chapter 4 : โ Mean (p. 172) โ Variance (p. 174) % %ยฃ ยง !ยจ ยฃ # ยง ยค ยฃ ยจ ยง ยค ยฃ &"ยฉยง % %!ยจ ' ยค ยข ยค ยข ยฃ ยก ยฃ ยฃ โ Put the pieces together! ยฃ โ โ Probability Distribution (p. 169) !ยค ยจ ยฃ "ยจ ยฉ ยจ ยฃ ยค ยข ยก ยฃ ยจ ยค ยข ยงยก ยจยก ยฃ ยข ยงยก ยฃ ยจ ยก ยจ ยค ยข ยก ยฆ ยข ยก ยจ ยจ ยจ ยฉ ยข ยข ยก ยจ ยข ยงยก ยจ ยฃ ยก ยฃ # ยข ยงยก ยฃ ยจ ยฉยก ยข ยจยฃ โ mutually exclusive Discrete Random Variables (p. 166) # ยฃ \$ยฃ ยข ยก ยฉยฃ โ Continuous Random Variables (p. 166) ยก ยค ยข ยก ยฉยฃ โ Random Variable (p. 164) ยก . What is โ HIV Example(like 3.109, p. 158) โ ยฉยก ยข ยฃ ? and ยก Know (2) Independent (p. 131, 133): DW ยจ ยงยก ยฃ ยฅ ยฉยก ยข ยฃ ยฅ ยข ยฉยก ยฅ ยฃยก ยข ยก RW and (1) # ยข ยฃ ยก ยจ ยข ยข ยฃ ยจ ยก ยฃ ยจ ยฃ , (p. 101) ยข # ยข ยงยก ยฃ ยจ ยจยฃ ยจ # ยข ยงยก ยฃ ยค # ยฅยก ยฃ Chapter 3 : Mutually exclusive (disjoint) (p. 118) Complement (p. 115) 6 โ The standard deviation of (p. 174). is 8 ยก STA 2023 c B.Presnell & D.Wackerly - Review for Exam I 9 ยก Characteristics of a Binomial Random Variable (p. 179) # ยฃยจ โ identical trials or โ โ on each trial stays same trial to trial โ Trials are independent ยจยฃ ยจ (p. 183) for ยก # # # ยก ยก ยจ ยฃ number of trials, โs in # ยจ ยขยข# ยจ ยฃยก ยฃ If trials number of โ Variable of interest: : ยจยฃ ยจ ยงยฉ# ยฆ ยงยฃ ยค ยฅยจ ยฃ # ยก ...
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https://studylib.net/doc/7552693/thermodynamics-exam-1-info-problems | 1,702,024,099,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100739.50/warc/CC-MAIN-20231208081124-20231208111124-00602.warc.gz | 618,041,101 | 17,812 | # Thermodynamics Exam 1 Info/Problems
Thermodynamics Exam 1 Practice Problems
Dr Colton, Spring 2006
Sample Conceptual Questions (answer and explain):
1.
Is it possible for two objects to be in thermal equilibrium if they are not in thermal contact with each other?
2.
Chimneys are never used as a weight-bearing part of the structure of a building. Why?
3.
Do the Celsius and Fahrenheit temperature scales ever read the same?
4.
Explain why, when a thermometer is placed in hot water, the column of mercury first descends slightly and then rises.
5.
The pendulum of a grandfather clock is made of brass. When the temperature increases, does the clock increase, decrease, or remain the same?
6.
Why does a city have a higher temperature at night than does the surrounding countryside?
7.
Do hotter objects always contain more heat than cooler objects?
8.
If water is a poor thermal conductor, why can its temperature be raised quickly when it is placed over a flame?
9.
Why can you grab a hot wooden object and not get burned, whereas if you grabbed a hot metal object at the same temperature it would burn you?
10.
What does the ideal gas law predict about the volume of a sample of gas at absolute zero?
Why is this prediction incorrect?
11.
In which process is there a greater change in entropy, an isothermal or an adiabatic?
12.
What is
S for an irreversible change in which both endpoints coincidentally lie on an adiabat?
13.
Electrical energy can be converted to heat energy with an efficiency of 100%, by sending a current through a resistor and using “Joule heating”: P = I 2
R. Explain how this doesn’t violate the 2 nd
Law, which told us that the efficiency of an engine can never by 100%.
14.
What is the State Postulate?
15.
Consider an alcohol and a mercury thermometer that read exactly 0
C at the ice point and
100
C at the steam point. The distance between the two points is divided into 100 equal parts in both thermometers. Do you think they will give exactly the same reading at a temperature of, say, 60
C?
Thermo Exam 1 Practice Problems – pg 1
16.
What is the difference between intensive and extensive properties of a system.
17.
Name several intensive properties.
18.
A metal ball can pass through a metal ring. When the ball is heated, however, it gets stuck in the ring. What would happen if the ring, rather than the ball, were heated?
19.
What would happen if you heated up a “bimetallic strip”? (That’s a long, narrow, piece of metal that is made up of two different metals joined back-to-back.)
20.
Explain why the apparent expansion of a liquid in a glass bulb does not give the true expansion of the liquid. Would the apparent expansion be greater than or less than, the true expansion?
21.
Give an example of a process in which no heat is transferred to or from a system, but the temperature of the system changes.
22.
A block of wood and a block of metal are at the same temperature. When the blocks feel cold, the metal feels colder than the wood; when the blocks are hot, the metal feels hotter than the wood. Is there ever a temperature when the two blocks feel equally cold or hot to you?
23.
Why should thicker insulation be used in an attic than in the walls of a house?
24.
Is ice always at 0
C? Can it be colder? Can it be warmer?
25.
Why are double-paned windows better than single-paned windows?
26.
Does adding heat to a system always increase its internal energy?
27.
If the pressure and volume of a system are given, is the temperature always uniquely determined?
28.
Does a gas do any work when it expands adiabatically? If so, what is the source of the energy needed to do this work?
29.
Why does stainless steel cookware often have a layer of copper or aluminum at the bottom?
30.
Why do actual gases tend to obey the ideal gas law, when they are at low densities?
31.
Suppose two equal-sized rooms are connected via an open doorway, but one room is warmer than the other. How is this possible? Which room would have more air molecules?
32.
How might you keep a gas at a constant temperature in a thermodynamic process?
Thermo Exam 1 Practice Problems – pg 2
33.
Can a kitchen be cooled by leaving the door of an electric refrigerator open?
34.
If you clean your room, you have lowered the disorder in your house. How does this not violate the 2 nd
Law of Thermodynamics?
Sample Problems (answer and show work):
1.
You stick a spherical He balloon in the freezer. What happens to the size of the balloon? Be as quantitative as possible.
2.
You have a 1.000 cm diameter steel ball which you desire to pass through a 0.999 cm inner diameter Al ring. (a) If you heat them up both together, how hot must they get before you can accomplish your task. (b) If you heat up only the ring, how hot must it get? (You may look up the coefficient of expansion.)
3.
A grandfather clock’s pendulum is made from steel, and calibrated at 5
C such that one period is 1 second. How much time does it lose in a day, if the temperature increases to 35
C? (Recall that T
2
L
is the equation for the period of a pendulum.) g
4.
You have 2 gas tanks connected to each other by a valve. Tank A is twice the size of tank B.
They are both initially at 20
C. (a) If tank A starts at 10
4
Pa and tank B starts at 10
5
Pa, what is the final temperature and pressure of each tank when the valve is opened? (You may assume the ideal gas law, and that no heat is added nor work done.)
5.
A textbook gives the coefficient of volume expansion for air as 3.67
10
-3
near room temperature. Use the ideal gas law to obtain a theoretical value for this number, and compare.
6.
A dead-weight piston is set up with an ideal gas inside the chamber. The piston is made out of 50 cm thick copper. (Recall that there is atmospheric air above the piston pushing down on it.) If the system is in equilibrium with T = 1500K and contains 10 moles of gas, determine the volume of the gas.
7.
You add 50 g of ice at –5
C to 200 g of water at 25
C. What is the final temperature of the mixture, assuming that no heat is lost to the outside?
8.
You add 50 g of steam at 150
C to 50 g of ice at –5
C. What is the final temperature of the mixture, assuming that no heat is lost to the outside?
9.
A cylindrical copper rod of length 1.2 m and cross-sectional area of 4.8 cm
2
is insulated to prevent heat loss through its surface. The ends are maintained at a temperature difference of
100
C by having one end in a water-ice mixture and the other in a boiling water/steam
Thermo Exam 1 Practice Problems – pg 3
mixture. If you initially have 10 g of ice on the one end and lots of steam on the other, how long does it take to melt the ice?
10.
How long would it take to melt the ice in the previous problem if the rod were 0.6 m of Cu and 0.6 m of iron?
11.
How about if the rod in problem 12 were divided in half the other way: 1.2 m of 2.4 cm
2
Cu side by side with 1.2 m of 2.4 cm
2
iron?
12.
A “solar cooker” consists of a curved parabolic-like reflector which focuses sunlight onto the object to be heated. The solar power per unit area reaching the Earth at the location of a 0.50 m diameter solar cooker is 600 W/m
2
. Assuming that 50% of the incident energy is converted to heat energy, how long would it take to boil 1.0 L of water initially at 20
C?
13.
A 1 kg cube of steel is heated from 20
C until the volume expands by 0.10 %. (a) What is its final temperature? (b) How much work was done in the expansion by pushing against the atmosphere? (c) How much work was done in the expansion by having to raise its center of mass? (d) How much heat was added in the process?
14.
The tungsten filament of a certain 100W light bulb radiates 2W of light (the other 98 W of energy is carried away by convection and conduction). The filament has a surface area of
0.25 mm 2 and an emissivity of 0.95. How hot is the filament?
15.
At high noon, the Sun delivers about 1000 W/m
2
of radiant energy (or perhaps a bit less, in
Wisconsin!). Suppose this strikes a blacktop which is insulated from the ground below. (Or equivalently, suppose the dirt below the blacktop has a very low thermal conductivity.)
Ignoring conduction and convection, what temperature would you predict the blacktop to reach?
16.
A large cold object is at 273 K, and a large hot object is at 373 K. 8.00 J of heat energy is transferred from the hot to the cold object, which is not enough to substantially change their temperatures. What is the total entropy change of this process?
17.
Fifteen identical particles have various speeds: one has a speed of 2.00 m/s; two have speeds of 3.00 m/s; three have speeds of 5.00 m/s; four have speeds of 7.00 m/s; three have speeds of 9.00 m/s; and two have speeds of 12.0 m/s. (a) Find the average speed, the rms speed, and the most probable speed. (b) If the particles have a molar mass of 1000 g/mole, make a reasonable guess as to the likely temperature of the particles.
18.
A refrigerator with C.O.P. = 4.7, extracts heat from the inside at a rate of 250 J per cycle. (a)
How much work per cycle is required to operate the refrigerator? (b) How much heat per cycle is discharged into the room?
19.
An engine using a polyatomic (atoms not in a row) ideal gas is driven by this cycle: from A to B, the pressure increase to 3 times its original pressure while keeping V constant; from B to C, it expands adiabatically until it reaches 4 times the original volume; from C to D, the
Thermo Exam 1 Practice Problems – pg 4
pressure drops at constant V; from D to A it contracts adiabatically. (a) Sketch the cycle, indicating P, V, and T, for all points. (b) What is the efficiency of this cycle? (c) What’s the maximum efficiency possible between the high and low temperatures. (Leave all answers in terms of the original P, V, and T.)
20.
A monatomic ideal gas (1 mole) undergoes this cycle: starting from 1 atm, 0
C, it contracts adiabatically to P = 2 atm, then it expands isothermally, then it contracts isobarically. (a)
Sketch the cycle, indicating P, V, and T, for all points. (b) What is the efficiency of this cycle? (c) What is
S for each leg of the cycle?
21.
A monatomic ideal gas ( n moles) undergoes this cycle: (1) starting at V
1
, T
1
, it increases the temperature at constant volume to 3 T
1
; (2) from V
1
, 3 T
1
, it increases the volume at constant temperature to 2 V
1
; (3) from 2 V
1
, 3 T
1
, it decreases the temperature at constant volume back to the original temperature, T
1
; (4) from 2 V
1
, T
1
, it decreases the volume back to the original volume, V
1
. (a) Sketch the cycle on a P-V diagram. (b) In terms of n and T
1
, what is the net work done by the game per cycle? (c) What is the efficiency of the cycle?
Thermo Exam 1 Practice Problems – pg 5
Conceptual
1.
Yes, if they are both at the same temperature. Then they are each in thermal equilibrium with the same (imaginary) thermometer, and by the 0 th Law, they are in equilibrium with each other. Put another way, no heat would flow even if they were connected.
2.
The chimney is going to expand when it’s heated. Other parts of the building are not going to be expanding as much, and thus the chimney would cause a stress on those parts if it were rigidly attached in a load-bearing manner.
3.
Yes, at –40 degrees. Solve T(F) = 9/5 T(C) + 32 for T(F) = T(C)
4.
The glass expands first, and then the mercury.
5.
The brass will expand, and have a larger period. The clock will therefore run slow.
6.
Concrete has a higher specific heat than does soil, so more heat is stored up during the day.
This heat can then be released at night.
7.
The heat “contained” by an object would most closely correlate with Q = mc
T; so a object with a smaller temperature could contain more heat if it had a larger “thermal mass” (mass
specific heat).
8.
Convection currents in the water transfer the heat.
9.
Wood has a much lower thermal conductivity, so the heat cannot be transferred to your fingers as quickly as it can be from the metal.
10.
The prediction would be for the volume to be 0. However, as the volume gets smaller & smaller, at some point the molecules start interacting with each other (which might, for example, cause the gas to condense into a liquid). At that point, the ideal gas law does not apply anymore because one of the basic assumptions in its derivation has been violated.
11.
Isothermal (assuming the volume is increasing); in an adiabatic change (no heat added),
S =
0.
12.
S = 0; since entropy is a state variable, it doesn’t matter how one goes from state A to state
B, the entropy change will be the same. So even though you don’t follow the adiabat, if you end up on it, there must be no entropy change.
13.
The 2 nd
Law is talking about efficiency of converting heat to work, not the other way around!
14.
The State Postulate reads: the state of a system is uniquely determined by 2 independent intensive properties.
Thermo Exam 1 Practice Problems – pg 6
15.
Probably, but not necessarily. They will if both liquids expand linearly with temperature, which is often the case.
16.
Intensive properties are those which do not depend on how much of the material you consider. Extensive properties, on the other hand, do depend on the quantity of material.
17.
temperature, pressure, density, entropy per mass, internal energy per mass.
18.
The ball would likely pass through easily.
19.
The strip would bend, because the two metals will expand at different rates.
20.
The glass is also expanding, which would cause it to look like the liquid is expanding less than it actually is.
21.
Any time you force a gas to expand or be compressed, you’re likely to change the temperature of the gas.
22.
They would both feel the same when they are at the same temperature as your skin—then no heat would flow and neither one would feel either hot or cold.
23.
Hot air rises; therefore all things being equal, there would be substantially more heat transfer from the attic to a cold outside than from the interior walls.
24.
Ice can obviously be colder than 0
C; if you take an ice cube an put it outside on a cold La
Crosse winter day, it will probably become much colder than 0
C. Ice may be able to exist at temperatures above that, but certainly not at atmospheric pressure (and probably not much warmer, in any event).
25.
The air in the middle of the two panes acts as an insulator, adding to your overall “R” value.
26.
No; for example, you could conceivably cause a system to do work at a constant temperature, in which case there is no change to the internal energy.
27.
Only if you know how much material you have (eg. how many moles or how much mass).
28.
Sure; when a gas expands it almost always does work (the “free expansion” was one exception). In this case, the source of energy would be from the internal energy of the molecules.
29.
As mentioned in class, stainless steel is a rather poor conductor of heat. If the bottom of the pan were made of stainless steel, it would take a long time for the heat to be conducted to your food.
Thermo Exam 1 Practice Problems – pg 7
30.
At low densities, the molecules of the gas are far apart, and so they do not interact with each other very much. That was one of the assumptions behind the derivation of the ideal gas law via “kinetic theory”.
31.
One room might be next to a warm stove; the other next to the cold outside. The colder room would have denser air, and hence more air molecule for the same volume.
32.
You’d have to add heat at precisely the same rate that the gas is doing work.
33.
No, that would actually warm the kitchen, unless the back of the refrigerator (where the heatexchanging coils are) were shoved out the wall. (In that case, the fridge would be acting as a heat pump.)
34.
It took energy to clean the room. The energy presumably came from some work converted from heat, which also involved some exhaust heat. The entropy added to the universe from that process more than compensates for the entropy decrease in your room.
Thermo Exam 1 Practice Problems – pg 8
Problems
1.
Diameter decreases ~ 3.5%
2.
T = 77.1
C; 41.7
C
3.
14.3 s
4.
20
C, 40 kPa
5.
For T
20
,
0.00341 /
C—pretty close!
6.
0.861 m
3
7.
3.59
C
8.
100
C
9.
210 s
10.
628 s
11.
349 s
12.
12.2 hrs
13.
50.3
C, 0.017 J, 8.2
10
-5
J, 1.36
10
4
J
14.
3490 K
15.
16.
0.0079 J/K
17.
(a) v ave
= 6.80 m/s, v rms
= 7.41 m/s, v mp
= 7.00 (b) based on the average and rms speeds, the temperature is likely around 2.2 K.
18.
53 J, 303 J
19.
(a) P, V, T; (b) 3 P, V, 3 T; (c) 0.4725 P,
4 V, 1.89 T; (d) 0.1575 P, 4 V, 0.63 T;
37%; 79%
20.
(a) 1.01
10
5
, 0.2246, 273; (b) 2.02
10
5
, 0.01482, 360.2; (c) 1.01
10
5
,
0.02964, 360.2; 12.7 %; 0, 5.76, –5.76
21.
(a) see Fig P22.57 on pg 701 in book, (b)
2 nRT
1
ln2, (c) 27.29%.
Thermo Exam 1 Practice Problems – pg 9 | 4,396 | 17,080 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2023-50 | latest | en | 0.923242 |
https://www.studysmarter.co.uk/explanations/math/geometry/area-of-a-kite/ | 1,719,003,114,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862157.88/warc/CC-MAIN-20240621191840-20240621221840-00636.warc.gz | 879,547,115 | 62,746 | # Area of a Kite
One day, Robert was flying his kite at the park when it suddenly got stuck in between some branches of a tree. When he finally managed to retrieve it, he found the plastic film on his kite ripped in the middle. To replace the body of his kite, he needs to find the area in order to purchase the correct amount of plastic film at the hardware store. Given that the frame of his kite provides the required dimensions of its diagonals, is there a particular formula he could use to determine the area?
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As a matter of fact, there is! In this article, we will discuss a formula that calculates the area of kites and observe several worked examples that employ this technique.
## Recap. Defining a Kite
Before we begin, let us begin by refreshing our memories of kites. A kite is a type of quadrilateral that has two pairs of equal adjacent sides. Like all other quadrilaterals, it contains 4 sides, 4 angles, and 2 diagonals.
The structure of a kite satisfies the characteristics of a cyclic quadrilateral. A cyclic quadrilateral is a quadrilateral where all four of its vertices lie on a circle. It is sometimes referred to as an inscribed quadrilateral. The circle that holds all four of these vertices on its circumference is called the circumcircle or circumscribed circle. Here is a diagram of a kite within a circle.
### Properties of a Kite
Let us now recall the fundamental properties of a kite. Here we have a kite denoted by ABCD. M is the point at which the diagonals intersect.
Diagram of a kite
The following table is a list of its features.
Properties of a Kite Description It has two pairs of equal adjacent sides AB = BC and AD = DC It has one pair of equal opposite angles that are obtuse ∠BAD = ∠BCD > 90o No parallel lines It has two non-equal diagonals AC ≠ BD Diagonals are perpendicular and bisect each other AC ⊥ BD and AM = MC and BM = MD
## The Area Formula for a Kite
The area of a kite is the space bounded by its sides. Referring back to our previous diagram of a kite, the area formula is given by
$A=\frac{1}{2}\times d_1 \times d_2$
where $$d_1$$ and $$d_2$$ are the lengths of the vertical diagonal and horizontal diagonal, respectively.
Area of a kite
### Deriving the Area of a Kite
Now we have an explicit recipe for finding the area of a kite. But how did it come about? This segment will discuss a step of step derivation of how this formula actually satisfies the area of a given kite. Again, let's turn our attention back to our previous kite, shown below.
Area of a kite
For our kite ABCD above, let's call the length of the shorter diagonal $$AC=x$$ and the length of the longer diagonal $$BD=y$$. From the properties of a kite, both these diagonals are perpendicular (at right angles) and bisect each other.
With that in mind, we have
$AM=MC=\frac{AC}{2}=\frac{x}{2}$
The area of kite ABCD is made up of the sum of two areas: triangle ABD and triangle BCD. Writing this as an expression, we have
Area of kite ABCD = Area of ΔABD + Area of ΔBCD
Let's called this Equation 1.
The area of a triangle is the product of its base and height multiplied by half, that is,
$\text{Area of a Triangle}=\frac{1}{2}\times b \times h$
where $$b$$ is the base and $$h$$ is the height. Using this formula, let's determine the areas of triangle ABD and triangle BCD.
$\text{Area of Triangle ABD}=\frac{1}{2}\times AM\times BD$
$\text{Area of Triangle BCD}=\frac{1}{2}\times MC\times BD$
Now replacing AM, BD and MC by $$x$$ and $$y$$, we have
$\text{Area of Triangle ABD}=\frac{1}{2}\times\frac{x}{2}\times y=\frac{xy}{4}$
$\text{Area of Triangle BCD}=\frac{1}{2}\times\frac{x}{2}\times y=\frac{xy}{4}$
Now, using Equation 1, we obtain
$\text{Area of kite ABCD}=\frac{xy}{4}+\frac{xy}{4}=\frac{xy}{2}$
Finally, substituting the values of $$x$$ and $$y$$, we have the required formula for the area of a kite.
$\text{Area of a kite}=\frac{1}{2}\times AC \times BD$
## The Area of a Kite and a Rhombus
The area formula of a kite happens to follow the same idea as the area of a rhombus. Let's recall the structure of a rhombus. Here we have a rhombus denoted by ABCD. M is the point at which the diagonals intersect.
Diagram of a rhombus
You can already see the resemblances with a kite, just by looking at this diagram. The following table is a list of its features.
Properties of a Rhombus Description Has four equal sides AB = BC = CD = DA Has opposite angles of equal measures ∠ABC = ∠CDA and ∠BCD = ∠DAB Has two pairs of parallel sides AB // DC and AD // BC Has two non-equal diagonals AC ≠ BD Diagonals are perpendicular and bisect each other AC ⊥ BD and AM = MC and BM = MD
The Area Formula for a Rhombus
$A=\frac{1}{2}\times d_1 \times d_2$
where d1 and d2 are the lengths of the vertical diagonal and horizontal diagonal, respectively.
Area of a rhombus
## Examples of Area of Kites
In this section, we shall look at several worked examples that make use of this formula that deduces the area of a kite. Here is the first example.
Cathy has 3 identical kite-shaped notecards with diagonals of lengths 5 inches and 17 inches. Determine the sum of the area for these 3 notecards.
Solution
The diagonals of each box are given by $$d_1=5$$ and $$d_2=17$$. Using the area formula for a kite, the area of one notecard is
$A=\frac{1}{2}\times 5 \times 17=\frac{82}{2}=42.5$
Thus, the area of each kite is 42.5 in2. Since we have 3 identical notecards, we can simply multiply this area by 3 to find their total area.
$42.5\times 3=127.5$
Thus, the total area of all 3 notecards is 127.5 in2.
Let's look at another example.
Mary has a cardboard cutout shaped like a kite. The shorter diagonal measures 3 feet while the longer diagonal measured is 14 feet. What is the area of this cutout?
She then decides to divide this cutout into 7 separate pieces of equal areas. What would the area of each piece be?
Solution
The diagonals of this cutout are given by $$d_1=3$$ and $$d_2=14$$. Using the area formula for a kite, the area of this cutout is
$A=\frac{1}{2}\times 3 \times 14=\frac{42}{2}=21$
Thus, the area of this cutout is 21 ft2. Since Mary wants to divide this cutout into 7 identical segments, we can simply divide this area by 7 to identify the area of each piece.
$\frac{21}{7}=3$
Hence, the area of each piece would be 3 ft2.
Here is one last example before we end this topic.
David has a kite with an area of 304 square inches. The shorter diagonal is 16 inches long. What is the length of the longer diagonal?
Solution
In this question, we are given the measures of the area and one of the diagonals of this kite, namely $$A=304$$ and $$d_1=16$$. In order to find the length of the longer diagonal, $$d_2$$, we need to rearrange the given formula to make $$d_2$$ the subject. Given that the formula for the area of a kite is
$A=\frac{1}{2}\times d_1 \times d_2$
Rearranging this formula so that $$d_2$$ becomes the subject yields
$d_2=\frac{2A}{d_1}$
Now substituting our known values for $$A$$ and $$d_1$$, we have
$d_2=\frac{2\times 304}{16}=38$
Thus, the length of the longer diagonal is 38 inches.
## Area of Kites - Key takeaways
• A kite is a type of quadrilateral with no parallel lines.
• A kite has two pairs of equal adjacent sides and one pair of equal opposite angles that are obtuse.
• A kite has two non-equal diagonals.
• The diagonals of a kite are perpendicular and bisect each other.
• The area of a kite is given by $A=\frac{1}{2}\times d_1 \times d_2$ where $$d_1$$ and $$d_2$$ are the lengths of the vertical diagonal and horizontal diagonal, respectively.
#### Flashcards in Area of a Kite 15
###### Learn with 15 Area of a Kite flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
How to find the area of a kite?
The area of a kite is the product of both its diagonals multiplied by half.
What is the area of a kite?
The area of a kite is the product of both its diagonals multiplied by half.
What are the properties of a kite?
• A kite is a type of quadrilateral.
• A kite has two pairs of equal adjacent sides.
• A kite has one pair of equal opposite angles that are obtuse.
• A kite has no parallel lines.
• A kite has two non-equal diagonals.
• The diagonals of a kite are perpendicular and bisect each other.
Are diagonals of a kite congruent?
No
Are the diagonals of a kite perpendicular?
Yes
## Test your knowledge with multiple choice flashcards
A kite has parallel lines.Is this statement True or False?
A kite has two pairs of equal adjacent sides.Is this statement True or False?
A kite has one pair of equal opposite angles that are acute.Is this statement True or False?
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• Smart Note-Taking | 2,559 | 9,977 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.875 | 5 | CC-MAIN-2024-26 | latest | en | 0.950564 |
https://www.ilovefreesoftware.com/21/windows/productivity/finance/free-excel-based-debt-reduction-calculator-to-payoff-credit-card-debt.html | 1,701,637,507,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100508.53/warc/CC-MAIN-20231203193127-20231203223127-00776.warc.gz | 943,465,277 | 26,159 | Editor Ratings:
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Works with: Microsoft Office, OpenOffice
To start with this Free Excel based Debt Reduction Calculator, you first enter all your debts in this excel. You can specify all kinds of debts in this: Credit Card Debt, Auto loan, Student loan, personal loan, and any other debt. After that, you specify interest rate of each debt and minimum payment required to be paid for each. This is the minimum payment that you financial institution has stipulated for your debt.
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This Free Excel based Debt Reduction Calculator is very well designed, and is extremely easy to use. More importantly, it helps you in getting rid of your debts, which is great. Love it!
Vertex42 has an interesting free Excel based Debt reduction calculator. This excel lets you determine the best strategy to payoff all your debts, and calculate the exact payments to be paid, and which debt to be paid off first. This Excel based Debt Reduction Calculator is totally free. You do not need to install anything, and there is no macro to run.
To start with this Free Excel based Debt Reduction Calculator, you first enter all your debts in this excel. You can specify all kinds of debts in this: Credit Card Debt, Auto loan, Student loan, personal loan, and any other debt. After that, you specify interest rate of each debt and minimum payment required to be paid for each. This is the minimum payment that you financial institution has stipulated for your debt.
Once you have entered this info, you need to enter the maximum monthly payment that you can make towards all of your debts combined. Once you have specified this, Free Excel based Debt Reduction Calculator will calculate the “snowball”, or extra payment by finding a difference between your total minimum payments that you must pay each month, and this maximum amount that you can pay each month. For example, if you have one credit card debt with minimum monthly payment of \$100, and another debt with minimum monthly payment of \$50, then total minimum monthly payments for month are \$150. If you can actually make payments of \$175 per month, then your initial snowball will be \$25.
Based on this extra payment, Free Excel based Debt Reduction Calculator will give you option to reduce your debt based on various debt reduction strategies.
#### Here are some of the strategiess available for Debt reduction in this Free Excel based Debt Reduction Calculator:
1. Debt Snowball (Lowest Balance First): This strategy aims to boost your morale by quickly paying off some of the debts. It starts with the debts that have lowest balance, and aims at paying that off first. This is good for those who have too many debts. This strategy would help them in getting rid of smallest debts first, and then focus on the bigger ones.
2. Highest Interest First: Financially speaking, this is actually the best strategy for Debt Reduction. It aims at paying off those loans first that carry highest interest rates, like credit card debts. Your total outflow of money will be lowest in this strategy, simply because you keep paying highest interest loans first.
3. User Specified Order: You can also choose to decide yourself about the order in which you need to pay off your debts.
Once you have decided your strategy for Debt reduction, this Free Excel based Debt Reduction Calculator will show you the exact payments that you need to make for each debt in each month to eliminate your debts. You can experiment between various Debt reduction strategies to see the payment plan for each, and see what works best for you. This also shows a Snowball Growth Chart to show how your interst dues decreases over time.
This Free Excel based Debt Reduction Calculator is very well designed, and is extremely easy to use. More importantly, it helps you in getting rid of your debts, which is great. This works with both Microsoft Office, as well as OpenOffice. Love it! | 1,207 | 6,047 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2023-50 | longest | en | 0.941395 |
https://www.shaalaa.com/question-bank-solutions/verify-property-x-y-z-x-y-x-z-taking-x-3-4-y-5-2-z-7-6-properties-rational-numbers-distributivity-of-multiplication-over-addition-for-rational_59897 | 1,618,895,027,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039375537.73/warc/CC-MAIN-20210420025739-20210420055739-00058.warc.gz | 1,096,988,930 | 9,334 | # Verify the Property: X × (Y + Z) = X × Y + X × Z by Taking: X = − 3 4 , Y = − 5 2 , Z = 7 6 - Mathematics
Sum
Verify the property: x × (y + z) = x × y + x × z by taking:
$x = \frac{- 3}{4}, y = \frac{- 5}{2}, z = \frac{7}{6}$
#### Solution
$\text{We have to verify that} x \times (y + z) = x \times y + x \times z .$
$x = \frac{- 3}{4}, y = \frac{- 5}{2}, z = \frac{7}{6}$
$x \times (y + z) = \frac{- 3}{4} \times (\frac{- 5}{2} + \frac{7}{6}) = \frac{- 3}{4} \times \frac{- 15 + 7}{6} = \frac{- 3}{4} \times \frac{- 8}{6} = 1$
$x \times y + x \times z = \frac{- 3}{4} \times \frac{- 5}{2} + \frac{- 3}{4} \times \frac{7}{6}$
$= \frac{15}{8} + \frac{- 7}{8}$
$= \frac{15 - 7}{8}$
$= 1$
$\therefore \frac{- 3}{4} \times (\frac{- 5}{2} + \frac{7}{6}) = \frac{- 3}{4} \times \frac{- 5}{2} + \frac{- 3}{4} \times \frac{7}{6}$
$\text{Hence verified .}$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 8 Maths
Chapter 1 Rational Numbers
Exercise 1.6 | Q 3.4 | Page 32 | 459 | 1,003 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2021-17 | latest | en | 0.408302 |
https://www.quesask.com/ | 1,638,107,245,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358560.75/warc/CC-MAIN-20211128134516-20211128164516-00188.warc.gz | 1,057,735,453 | 22,671 | ### Geetika saves 25 per day. Find the least number of days in which she would be able to sa money in
Geetika saves 25 per day. Find the least number of days in which she would be able to sa
money in multiples of 5002
### 1. Aye aye aye 2. Siren head3. Industry baby 4. Let her go Follow for more songs
1. Aye aye aye 2. Siren head3. Industry baby 4. Let her go Follow for more songs
### On the basis of kinetic model, explain how the liquid exerts pressure short
On the basis of kinetic model, explain how the liquid exerts pressure short answer
### In tissue of vertebrate fixed for FormalineAldehydeCarnoy”s fluidBouin”s flui
In tissue of vertebrate fixed for
FormalineAldehydeCarnoy”s fluidBouin”s flui
### 1.2. Choose the correct pair – a. Upper course of a river – gentle slope of a land b. Upper course
1.2. Choose the correct pair –
a. Upper course of a river – gentle slope of a land b. Upper course of a river – dominant work is erosion c. Lower course of a river – steep stope of a landd. Lower course of a river – dominant work is transportation
### If x-y = 2 , then the volue of x³-y³ -6xy is (a) 7 (b) 8 (c) 9 (d) 10Give me
If x-y = 2 , then the volue of x³-y³ -6xy is
(a) 7 (b) 8 (c) 9 (d) 10Give me the write answer
### If Assertion two matrik A and B will multiply each only if they are of unequal order Reason— order in
if Assertion two matrik A and B will multiply each only if they are of unequal order Reason— order in case of multiplication plays significant role
### A more reactive metal can replace a less reactive metal, but a less reactive I can not replace a more
a more reactive metal can replace a less reactive metal, but a less reactive I can not replace a more reactive metal. is this true or false.
### 3. The product of the fraction, fraction and the difference of 35 and at a la final 3
3. The product of the fraction, fraction and the difference of 35 and at a la final 3
### Which of the following is a polynomial? a. x²+(5√x)+7 b. (1/x³)+7c. 3x²+7 d. √x-1
Which of the following is a polynomial?
a. x²+(5√x)+7 b. (1/x³)+7c. 3x²+7 d. √x-1 | 631 | 2,114 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2021-49 | latest | en | 0.823533 |
https://twistedsifter.com/2024/05/what-is-time-dilation-and-what-will-it-do-to-astronauts-who-want-to-travel-to-mars/ | 1,716,477,820,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058642.50/warc/CC-MAIN-20240523142446-20240523172446-00557.warc.gz | 513,036,892 | 19,263 | May 5, 2024 at 12:38 pm
# What Is Time Dilation And What Will It Do To Astronauts Who Want To Travel To Mars?
There is probably an unbelievably long list of the challenges that human astronauts would face in the event of a manned mission to Mars.
Dealing with time dilation is definitely on the list.
We all experience time dilation to some degree. Time passes at different rates for different observers, depending on their relative speeds and their proximity to (and strength of) nearby gravitational fields.
Weird, right?
Time dilation is the difference between time elapsed on two clocks depending on their special relativity and general relativity.
Because of the way gravity curves spacetime, gravity is stronger near you, and the closer you move to the mass creating gravity, the slower time moves.
The further you get away from Earth’s gravity – like, if you’re on top of a tall building – the more pronounced the effect compared to those on the ground.
Not a lot in that example; a fraction of a nanosecond a year.
It is much more pronounced for astronauts experiencing zero-gravity for long periods of time.
That said, astronomer Colin Stuart says it’s almost canceled out by the speed at which they travel through space.
“Because astronauts and satellites orbiting the Earth are slightly further away from the centre of the planet (compared to people on the ground) they actually experience less gravitational time dilation. O its own this would mean astronauts’ time would run faster. However, this effect is quite small because Earth’s gravity is quite weak and so the time dilation due to their speed wins out and astronauts really do travel a tiny amount into the future.”
When Cosmonaut Sergei Krikalev was stranded in space for 311 days, he technically traveled 0.02 seconds into the future.
A trip to Mars would take around 21 months, roundtrip. The astronauts will experience time passing normally, but it will actually be a few nanoseconds difference from how we experience life on Earth.
If they stayed on Mars for extended periods of time, the effects of gravitational time dilation could be enough to be noticeable.
For example, if you lived 80 years on Mars, you would pass away 12 seconds earlier than if you passed away after exactly 80 years on Earth.
From the point-of-view of the person experiencing either end, though, it would feel exactly the same.
Spacetime is wild, y’all.
I have a feeling we’re going to have the opportunity to study it more in depth very soon. | 522 | 2,514 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2024-22 | latest | en | 0.952945 |
https://ctspedmathdude.com/tag/analyze-graph/ | 1,632,221,186,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057202.68/warc/CC-MAIN-20210921101319-20210921131319-00588.warc.gz | 254,418,457 | 18,943 | ## Analyzing a Graph
Students can hit a road block at the steps that appear to be very simple. For example, in the problem below the students are prompted to find the highest point on the graph. Many think the graph refers to the entire coordinate plane and they pick 5 as the high point. It is the highest point on the y-axis but not the graph. I introduce the problem by highlighting the actual graph in pink and explain that this highlighted line is what is meant by the graph.
The use of color also helps students distinguish between the x and y axes and what the variables x and y represent in the context of the problem (# minutes and # kilometers in this problem) – see photo above. This problem also involves plugging in a # for x (blue) IN the function (red). In the photo below you see how I use color to help emphasize this. | 181 | 837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.0625 | 3 | CC-MAIN-2021-39 | latest | en | 0.944815 |
https://becca9941.github.io/kata/2019/08/09/understanding-three-n-plus-one-problem.html | 1,585,648,670,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370500426.22/warc/CC-MAIN-20200331084941-20200331114941-00064.warc.gz | 375,369,955 | 3,821 | # Understanding 3n + 1 problem
09 Aug 2019
Start with an integer n. If n is even, divide by 2. If n is odd, multiply by 3 and add 1. Repeat this process with the new value of n, terminating when the value is 1. For an input n, the cycle length of n is the number of numbers generated up to and including the 1. The cycle length of 22 is 16. Given any two numbers i and j, you are to determine the maximum cycle length over all numbers between i and j, including both endpoints.
## Sample Input
``````
1 10
100 200
201 210
900 1000
```
```
## Sample Output
``````
1 10 20
100 200 125
201 210 89
900 1000 174
```
```
## Do I understand the problem?
Here are the key rules that stand out to me:
### Rules as I understand them
• If n is even, divide by two
• If odd, multiply by 3 then add 1
• Stop after you have reached the end point
• Keep track of how many cycles there are up to and including the end point.
The first thing I'm going to do is write a test list which includes all of the sample output cases, as well as the example mentioned in the problem statement.
### Test list
• 1 22 16
• 1 10 20
• 100 200 125
• 201 210 89
• 900 1000 174
The next thing I'm going to do is work out the example given in the problem statement by hand to see if it matches their solution
``````
22 is even so / 2
11 is odd so x 3 + 1
34 is even so / 2
17 is odd so x 3 + 1
52 is even so / 2
26 is even so / 2
13 is odd so x 3 + 1
40 is even so / 2
20 is even so / 2
10 is even so / 2
5 is odd so x 3 + 1
16 is even so / 2
8 is even so / 2
4 is even so / 2
2 is even so / 2
1
Cycle length = 16.
```
```
Okay, my workings out matched the example in the problem statement. Now I'm going to do the same thing for the first sample example in the test output, which is '1 10 20'
``````
10 is even so / 2
5 is odd so x 3 + 1
16 is even so / 2
8 is even so / 2
4 is even so / 2
2 is even so / 2
1
cycle length: 7
```
```
Well, something went wrong, the expected output was supposed to be 10. I've done the same thing as I did for the 22 example. They both had the same end point of 1.
Reading the problem again, it says determine the 'maximum' cycle length, which implies that there is a 'minimum' cycle length. The emphasis was on including both end points. Was the 1 included in the cycle for 22? Yes.
Hmm, maybe I need to do do the cycle for each individual number from 1 to 10. Let's try it
Cycle length of 10 is 7
9,28,14,7,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1
Cycle length of 9 is 20, which takes us over the 20 limit, so this isn't the way to go. OH WAIT, maybe that IS the answer. So are we counting the cycles individually for each number and keeping the largest score?? THAT MAKES SENSE.
I did not expect the first problem in the book to be this difficult. But okay, now I think I understand it properly.
## Rephrasing problem statement
My program then must count the cycles for each number in the range including both the start and end number, and then return the largest of all of those cycle counts.
Now that I understand the problem statement, the next step is to think about how I'm going to solve the problemas well as set up my developer environment. Will link to it here once the article is started.
Wow, it took me a good 2 hours to actually understand what this problem was actually asking, though I was also getting distracted by a kitten who keeps pouncing on me and the keyboard. He's currently lying on my chest trying to boop my nose with his paws, heart melted.
Time to get ready for work. | 1,026 | 3,528 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2020-16 | longest | en | 0.944984 |
https://theharbourleague.org/sum-and-difference-of-cubes-worksheet/ | 1,656,320,486,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103329963.19/warc/CC-MAIN-20220627073417-20220627103417-00630.warc.gz | 597,923,077 | 9,577 | HomeTemplate ideas ➟ 0 Sum And Difference Of Cubes Worksheet
# Sum And Difference Of Cubes Worksheet
Sum And Difference Of Cubes Worksheet. When factoring difference and sum of two terms worksheet for each problem? Factoring sum and difference of two cubes:
3 + 81{x^3}{y^3} problem 5: Today we will factor 2 terms binomials and 4 polynomials terms difference of. The sum of cubes, difference of cubes, or neither.
### Factoring The Sum Or Difference Of Cubes Worksheet.
Factor out each binomial completely. Factoring the sum or difference of cubes worksheet. Next, multiply the terms of the binomial factor to create the middle term of the trinomial factor.
### Example Of Cash Is And Sum And The Following.
Sum of all three digit numbers formed using 1, 3, 4. Factoring sum and difference of two cubes: Since so many students have a difficult time remembering the rules for the sum.
### This Terrific Activity For Factoring Binomials.
Get the cube root of the 1st term, then the 2nd term to get the binomial factor. | 242 | 1,026 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2022-27 | latest | en | 0.768135 |
http://www.solutioninn.com/in-exercise-24-an-estimated-regression-equation-was-developed-relating | 1,508,344,389,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823016.53/warc/CC-MAIN-20171018161655-20171018181655-00683.warc.gz | 566,832,476 | 8,128 | In exercise 24 an estimated regression equation was developed relating
In exercise 24, an estimated regression equation was developed relating the percentage of games won by a team in the National Football League for the 2011 season (y) given the average number of passing yards obtained per game on offense (x1) and the average number of yards given up per game on defense (x2) (ESPN website, November 3, 2012). The estimated regression equation was
y = 60.5 + .319x1 – .241x2
a. Predict the percentage of games won for a particular team that averages 225 passing yards per game on offense and gives up an average of 300 yards per game on defense.
b. Develop a 95% confidence interval for the mean percentage of games won for all teams that average 225 passing yards per game on offense and gives up an average of 300 yards per game on defense.
Membership | 194 | 857 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2017-43 | latest | en | 0.959515 |
https://keisan.casio.com/exec/system/1272552070 | 1,553,611,179,000,000,000 | text/html | crawl-data/CC-MAIN-2019-13/segments/1552912205534.99/warc/CC-MAIN-20190326135436-20190326161436-00115.warc.gz | 524,628,885 | 8,706 | # Area of a triangle given sides and angle Calculator
## Calculates the area, perimeter and height of a triangle given two sides and the angle.
side a side b angle θ degreeradian 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit area S perimeter L height h
$\normal Triangle\ (a,b,\theta\rightarrow S)\\(1)\ area:\hspace{50} S={\large\frac{1}{2}}ab {\hfill{1}} \sin{\theta}\\(2)\ perimeter:\ L=a+b+\sqrt{a^2+b^2-2ab{ {\hfill{1}} \cos{\theta}}}\\\vspace{5}(3)\ height:\hspace{40} h=b\hspace{2}sin{\theta}\\$
Area of a triangle given sides and angle
[1-4] /4 Disp-Num5103050100200
[1] 2017/09/13 16:53 Male / 50 years old level / An engineer / Very /
Purpose of use
Just to check whether term sqrt(a^2+b^2-2*a*b*cos(theta)), in formula for "perimeter", may give correct answer (altrough I have already known that it can''t); for example a=b=2^32, theta =1/2^48, just 26 significant figgures, out of 50, due to cosine being too close to 1, better method: c=sqrt((a-b)^2+4*a*b*(sin(theta/2))^2), but nobody uses it, except Professor William Kahan, a few other people and me :-)
Comment/Request
Maybe you may use term that I have proposed, instead of "ordinary used" one...
[2] 2017/09/09 14:06 Female / Under 20 years old / High-school/ University/ Grad student / Very /
Purpose of use
Check answers to see if the java program I wrote works
[3] 2017/08/16 09:54 Female / Under 20 years old / Elementary school/ Junior high-school student / Not at All /
Purpose of use
School
Comment/Request
[4] 2014/07/26 22:21 Male / 30 years old level / An engineer / A little /
Purpose of use
time past
Comment/Request
i need
Sending completion
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# Akshit
### Michigan University
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I have a set of 3D points (x,y,z) , I want to take each point and calculate the sum of distances from all other points from this point and find the minimum of it , could you help me with codes?
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nästan 8 år ago | 1 answer | 0 | 1,254 | 4,458 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.15625 | 3 | CC-MAIN-2022-21 | latest | en | 0.752979 |
https://devforum.roblox.com/t/offset-on-bezier-help-wanted/2772889 | 1,725,734,820,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650898.24/warc/CC-MAIN-20240907162417-20240907192417-00784.warc.gz | 199,799,210 | 5,835 | # Offset on Bezier - help wanted
Hello guys, I’m trying to learn beziers. But most hard thing I met is offsetting N studs from some point.
Because bezier has variying lengths between 2 point every time, I tried to make code which will constantly check offset length until wanted offest with treshhold achieved
I tried to use this code, modified it a lot, but sadly, for now best what I have got is increse of distance from desired dot.
Can someone help me to fix my code, so it will be able to calculate approximate offset from dot on bezier?
``````local function CreateCurve(Dots, Mode, Time, Params)
if Mode == "Length" then
local BezierCurve = {}
Time = Time or 1
if Time <= 0 then error("INVALID TIME") end
local i = 0
while i <= Params do
local Value = i/Params * Time
local Lerps = table.clone(Dots)
if Time == math.huge then error("Inf asquared") end
while #Lerps > 1 do
for a = 1, #Lerps-1, 1 do
Lerps[a] = Vector3.new(Lerps[a].X * (1 - Value) + Lerps[a+1].X * Value, 0, Lerps[a].Z * (1 - Value) + Lerps[a+1].Z * Value)
end
Lerps[#Lerps] = nil
end
table.insert(BezierCurve, Lerps[1])
i += 1
end
local BezierLength = 0
for i = 1, #BezierCurve - 1, 1 do
BezierLength += (BezierCurve[i + 1] - BezierCurve[i]).Magnitude
end
return BezierLength
elseif Mode == "Dot" then
local BezierCurve = {}
local Lerps = table.clone(Dots)
while #Lerps > 1 do
for a = 1, #Lerps-1, 1 do
Lerps[a] = Vector3.new(Lerps[a].X * (1 - Time) + Lerps[a+1].X * Time, 0, Lerps[a].Z * (1 - Time) + Lerps[a+1].Z * Time)
end
Lerps[#Lerps] = nil
end
return Lerps[1]
elseif Mode == "Offset" then
local BezierSize = Params[1] or CreateCurve(Dots, "Length", nil, 100)
local StartLengthToDot = CreateCurve(Dots, "Length", Time, 25)
local BaseOffset = Params[2]
--local LengthToDot = StartLengthToDot
local Dot
local Breakage = 100
local CheckTime = Time
--local AverageLength = LengthToDot / Time
local Offset = BaseOffset
local Variativity = 1
warn(BezierSize)
--warn(LengthToDot)
repeat
local CurOffset = BaseOffset * Variativity
CheckTime = (StartLengthToDot + CurOffset) / BezierSize --0.4
local LengthToDot = CreateCurve(Dots, "Length", CheckTime, 25)
local Marker = Instance.new("Part")
Marker.Anchored = true
Marker.Size = Vector3.new(0.15, 0.15, 0.15)
Marker.Position = CreateCurve(Dots, "Dot", CheckTime)
Marker.Color = Color3.fromRGB(0, 255 * ((100 - Breakage) / 100), 255 * ((100 - Breakage) / 100))
Marker.Parent = workspace
Offset = LengthToDot - StartLengthToDot
Variativity = BaseOffset / Offset
warn(Variativity, CheckTime)
--warn("\nCheck time: " .. tostring(CheckTime) .. "\nLengthToDot is " .. tostring(LengthToDot) .. "\nCurOffset is " .. tostring(CurOffset) .. "\nOffset is " .. tostring(Offset))
Breakage -= 1
if Breakage <= 0 then
error("Too many iterations")
end
task.wait(0.5)
until math.abs(Offset) < 0.01
return CreateCurve(Dots, "Dot", CheckTime)
end
end
local Points = {
Vector3.new(0, 0, 0),
Vector3.new(50, 0, 50),
--Vector3.new(0, 0, 75),
--Vector3.new(-50, 0, -100),
Vector3.new(0, 0, 100),
}
for i = 1, 1000, 1 do
local Dot = CreateCurve(Points, "Dot", i/1000)
local Marker = Instance.new("Part")
Marker.Anchored = true
Marker.Size = Vector3.new(0.1, 0.1, 0.1)
Marker.Position = Dot
Marker.Color = Color3.fromRGB(255, 0, 0)
Marker.Parent = workspace
end
local Marker = Instance.new("Part")
Marker.Anchored = true
Marker.Size = Vector3.new(0.3, 0.3, 0.3)
Marker.Position = CreateCurve(Points, "Dot", 0.5)
Marker.Color = Color3.fromRGB(0, 0, 255)
Marker.Parent = workspace
Marker = Marker:Clone()
Marker.Position = CreateCurve(Points, "Offset", 0.5, {CreateCurve(Points, "Length", 1, 100), -10})
Marker.Color = Color3.fromRGB(0, 255, 0)
Marker.Parent = workspace
CreateCurve(Points, "Length", nil, 100)
`````` | 1,229 | 3,722 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2024-38 | latest | en | 0.557543 |
https://forums.kleientertainment.com/forums/topic/148420-would-this-be-a-somewhat-simple-and-guaranteed-way-to-domesticate-an-ornery-beefalo-every-time/ | 1,701,187,494,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679099892.46/warc/CC-MAIN-20231128151412-20231128181412-00372.warc.gz | 314,647,841 | 31,269 | # Would this be a somewhat-simple and guaranteed way to domesticate an Ornery beefalo every time?
## Recommended Posts
tldr; If the per-day-average number of hits you give/receive riding your beefalo are over 51, will you get the ornery beefalo tendency?
Math:
1) A day is 8 minutes = 480 seconds. Assume you spend every second riding everyday of the domestication process. Realistically this could never happen but it's a generous estimate. For every second you ride a beefalo you get +0.000416 to the rider tendency (source: wiki). That's 0.000416x480 = 0.2 points a day
2) Every second your beefalo is over 50% hunger (150+ hunger points), it gets +0.000416 added (same as rider) to its pudgy tendency. Let's assume you never feed your beefalo 150 hunger in one sitting. A diet that's mostly lightbulbs would accomplish this, and never two steamed twigs at once. Suppose you did go over for a while. We can assume the seconds you weren't riding were about the same as the error of overfeeding it. So roughly, you rack up a total of 0.2 points a day for both rider and pudgy combined. Conservative estimate
3) Every time your beefalo gets hit, or you hit a mob, you get +0.004 to the ornery tendency. In order to lock-in a tendency, DST says you need to have one tendency's points greater than the other two combined. This means you need more than 0.2 points a day. That comes out to 0.2 / 0.004 = 50 hits
Does means if you count an average of 51 hits per day over the course of your taming process, you would certainly get the ornery tendency? Theoretical estimation this, in practice you could get a safer estimate and do 60 hits per day. But my question is would this work? Or is there a console command or mod to test this? I installed the "Show me" mod but it doesnt show the tendency stat
you are correct.
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Ornery is like, the easiest tendency to get tho
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19 minutes ago, Shosuko said:
Ornery is like, the easiest tendency to get tho
I had no idea. I'm super new to taming beef so I remember players saying ornery was hard. Which is the hardest tendency? I got rider on my first time. And I assume pudgy should be easy/simple too.
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24 minutes ago, DarXide said:
I had no idea. I'm super new to taming beef so I remember players saying ornery was hard. Which is the hardest tendency? I got rider on my first time. And I assume pudgy should be easy/simple too.
Unless you do something to extend the time it takes to domesticate your beef (striking it to get +6 days timer) there is a maximum amount of time you can spend riding your beef, or keeping its hunger above the 50% threshold. These are naturally time gated. Attacking or being attacked, well I guess theoretically there is a limit but you can spend almost all day getting stung by a few bees, and there is no shortage of excuse to farm spiders, hounds, kill pigs etc early game. Its very easy to rack up a LOT of ornery domestication points even without realizing it, crowding out the other tendencies. So raising anything *not* ornery requires you don't use the beef for fighting much.
I wouldn't say any are "hard," but it is more restrictive b/c time spent off of it to fight means feeding it more and risking it shaking its saddle. The "hardest" is probably default because you have to balance the 3. If you don't do some of all three tendencies the threshold for it to be default is pretty thin and might take some work to get right.
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use the new punching bags
or if you're oldschool, carry a bunch of lureplants and keep replanting them
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4 hours ago, DarXide said:
I had no idea. I'm super new to taming beef so I remember players saying ornery was hard.
A huge number of players are extremely bad at this game. No need to listen to them. To get an ornery beefalo you can pick cacti*, fight spiders and bees* or carry around a Lure Plant. Having a 1-3 Lure Plants in your inventory makes dishing out damage on demand very trivial. Good luck taming your beefalo!
*) Taking damage also gives tendency points towards ornery.
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4 hours ago, Shosuko said:
Ornery is like, the easiest tendency to get tho
Idk pretty sure rider can be acquired by doing legitimately nothing but sitting afk on your beefalo and remounting it whenever it bucks you.
As for ornery beefalo I basically just ride them all the time, and clear out spider dens/bees since they do minimal damage and have very little health. Just be sure to wear a football helmet at all times, as getting bucked mid fight is quite common, and can often lead to grievous wounds or straight up death
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4 hours ago, DarXide said:
I had no idea. I'm super new to taming beef so I remember players saying ornery was hard
Just use the beefalo for combat and cactus picking, and if it's almost done domesticating and doesn't have a grumpy face start focusing hardcore on fighting. Or just smack a lureplant.
4 hours ago, DarXide said:
Which is the hardest tendency?
Pudgy, huge resource sink. Ornery & Rider can easily be obtained for essentially free by simply fighting with the beefalo or riding the beefalo. Pudgy requires you to feed him approximately 1.2 billion cubic tons of food per nanosecond (I did the math).
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8 minutes ago, Baark0 said:
Idk pretty sure rider can be acquired by doing legitimately nothing but sitting afk on your beefalo and remounting
You can turn any type of beefalo into ornery at the last day, but you cant do the opposite. It took half day (400 secs) to turn a 0 ornery point to 50% ornery point if you keep attack 2.5 times each sec. Even less if you poke killer bee hive with bunch of blue cap to heal beef back.
Especially in spring where you cant ride untamed beef into any danger, it better to just leave beef near salt lick and feed him steamed twigs, which wont give any rider points.
If you ask why people tame beef in spring, the reason is we join other pp server to play and cant decide when the season is right there.
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2 hours ago, Well-met said:
use the new punching bags
or if you're oldschool, carry a bunch of lureplants and keep replanting them
Does that really work?
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18 minutes ago, LanOrhan said:
Does that really work?
yes.
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ok you all seem to do the std taming stuff, maybe ill give you an advanced example ...i tamed 5 beefallo at once multiple times now, wilson and wicker working best for this bc you need a lot of twigs for this, you can make all 5 whatever you want them to be and you could even do more then 5.
Here is how:
Get 5 beefs(bells) into one spot(base/outpost) and i recommence groomingstations so they not all buzzing around all the time.
Drop the beefbell of the corresponding beef at their grooming stations and make a saltlick what covers all the beefs (saltlick is 1 pitchforktile larger radius then flingo so this one is easy.
Then you make 3 pots to cook steamed twigs and the setup is done.
Now the interesting bit, for the tendency domestication you just need to rider or fight about a day and it will have a tendency locked in after locking in tendency you can take the saddle off the beef and go to the next one.
After this you just need to feed those beefs 3 meals a day (i start mornings and then go by 1/3 of the clock) if you late dont worrie the saltlick will catch it the important thing is that you have at least 1/3 of the clock between meals so it dont aqire any fatty points.
now its just feeding and brushing (if brush availible) if you feed responsible it wont get faty points and it cant get any other points too since you tame without a saddle you cant ride it and since its on a station in base it wont fight either, so your tendency is final .
You can pause this process for a couple of days at any time since you got a saltlick, also other ppl can takeover for a couple of days in your place.
And for those who like the rider aproach (only feed lightbulbs be in cave every 3 days gathering new bulbs and ride all the time), well you have to work hard to override all that riding points either way fight all the time or gather healings and facetank killerbees for a while ..
In the end its preference ofc wich way one choose to tame beef but ill say a word about the 2 methods riding alltime vs. keep hunger >0 all time.
If you choose the riding way the drawback is obvious you HAVE TO be on your beef ALL the time, everytime you off beef you will lose points not just stop but strait out lose except if you feed him hunger gaining stuff asap and then you only have a few seconds if it wasnt steam twigs.So its advised to have some steamed twig on you anyways even if you roll with lightbulbs just for offbeef missions like chop wood pick stones etc.The upside ofc is that you get to ride your beef
For the feeding approach the drawback would be that you have to feed beefs every 1/3 of the clock for 20 days (default) what makes you kinda stationary, the upside would be that you have basicly no time to spend on beef and can play otherwise normaly doing all things and you even allowed slacking throu saltlick what will extend the time you need to tame thou.
For efficiency comperasent ....a beef what is starving will lose obidience at twice the rate as one who is not and that would be 50% nonstarving, so you will need to feed your beef 10 things a day if you riding approach it(more if you want ornery), so this better be bulbs and not twig or else you add 2 twigs and you have 3 steamed twigs.
My personal Recommendation is to pick the riding way if you want "beef gameplay" in wich case the char should be something synergysing like max with duelists for example.
If you dont wanna be beefplaying all the time and just want a fast car at the end of the process, well then i can only recommence trying the feeding approach, woodie would come to mind for example on this one cause this pick of char would be boarderline useless to make if you plan on riding all the time, woodie who cant transform (off the beef) and cant use lucy (off the beef) that would be idk ...bad choice XD
Hope i could help/inspire a bit here, if so thank klei for slacking... their servers are not updated to the new version jet so im bored
Have a nice day and gl with your tendencys even thou luck got nothing to do with it
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Foolproof method of getting/switching* to an Ornery tendency: (should have at least 40% domestication already so you don't get bucked off in the process)
1. Go to blue mushroom biome in caves, the one next to the lunar grotto. Pick or dig mushrooms.
2. Attack the tentapillar on a beefalo to summon the baby tentacles. 2 or 3 hits on the pillar will summon enough baby tentacles. Don't overwhelm the beef.
3. Tank the hits on the beef, each hit only does like 5 dmg and the beefalo has 1000 health. Be aware that your sanity will be tanking too since each tentacle has its own aura and a lot of them will be spawning on a small area.
4. Count 10s OR when the character comments on the beefalo's health(meaning its below 200). Move away then feed blue mushrooms to restore health, each heals the beef for 80 health.
5. Doing this 3 or 4 times is enough hits to guarantee the Ornery tendency. Take note that you will be insane after each session so mind the nightmares, they might kill the beef if you fight on it while its health is low.
*I switched to an ornery while having a rider beef at around 90% domestication cause a friend tamed a rider and they wanted ornery. I only use the beef for transportation cause I play wanda so it wasn't getting any ornery points outside the tentacle session.
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my way of getting ornery involves literally just going to the bee queen biome and committing killer bee genocide, the game does everything else from there
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On 6/16/2023 at 12:50 AM, DarXide said:
during the domestication process beefalo will change its face to reflect what tendency currently it has most points in
this can give you valuable clues to whether or not you're going the right path
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On 6/16/2023 at 3:30 AM, n11360 said:
Here is how:
This is also how I tame my beefalo nowadays. This way gives a lot more freedom, surprisingly. | 2,965 | 12,436 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2023-50 | latest | en | 0.947572 |
https://www.jiskha.com/display.cgi?id=1283889727 | 1,511,231,634,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806310.85/warc/CC-MAIN-20171121021058-20171121041058-00566.warc.gz | 832,807,865 | 4,086 | # Algebra
posted by .
A moving Company weighs 20 boxes you have packed that contain either books or clothes and says the total weight is 404 pounds. You know that a box of books weighs 40 pounds and a box of clothes weighs 7 pounds. Find how many boxes of books and how many boxes of clothes you packed.
• Algebra -
number of boxes of books ---> x
number of boxes of clothing ---> 20-x
solve
40x + 7(20-x) = 404
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More Similar Questions | 608 | 2,453 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.421875 | 3 | CC-MAIN-2017-47 | longest | en | 0.910618 |
http://nrich.maths.org/2305/solution | 1,477,378,151,000,000,000 | text/html | crawl-data/CC-MAIN-2016-44/segments/1476988719960.60/warc/CC-MAIN-20161020183839-00336-ip-10-171-6-4.ec2.internal.warc.gz | 177,083,388 | 5,998 | ### Rule of Three
If it takes four men one day to build a wall, how long does it take 60,000 men to build a similar wall?
### Crossing the Atlantic
Every day at noon a boat leaves Le Havre for New York while another boat leaves New York for Le Havre. The ocean crossing takes seven days. How many boats will each boat cross during their journey?
### On Time
On a clock the three hands - the second, minute and hour hands - are on the same axis. How often in a 24 hour day will the second hand be parallel to either of the two other hands?
# Buses
##### Stage: 3 Challenge Level:
This problem was more complicated than it seemed - we had over 50 solutions suggesting that when a bus travelled from one end to another it would meet buses travelling from the other direction every 10 minutes. Unfortunately this is not correct!
When the bus sets off there will be a bus arriving from the other end, a bus due to arrive 10 minutes later, a bus due to arrive 20 minutes later, a bus due to arrive 30 minutes later, and a bus just leaving at the other end which is due to arrive in 40 minutes. The bus will meet all these buses and all the buses that leave while it is travelling, that is, the buses that leave 10, 20 and 30 minutes after it has set off. It will arrive as another bus prepares to leave.
So the bus will set off as a bus arrives from the other end, meet a bus every 5 minutes while travelling (7 in all) and arrive at the other end as another is leaving.
If it is the first bus of the day, it will only meet 4 buses on its way, the second bus of the day will meet 5, the third bus will meet 6, the fourth bus will meet 7, the buses after the fourth one will all meet 7 except the last three - they will meet 6, 5 and 4 respectively.
Jessica from West Herts College sent a correct solution:
A bus in the middle of the day will either meet 7 or 9 depending on whether you count the buses it meets at either end of its journey. However the first and last bus will only meet 4 or 5 depending on whether you count the bus leaving as it arrives (in the case of the 1st bus) or leaves (in the case of the last bus).
Anna, PD and Emily from Ardingly College Junior School thought of it like this:
... if it left the station at 3.40pm it would meet the ones that left the station within 40 minutes before and after it and the one that left at the same time as it.
Heather from Stow Heath Junior School noticed that:
If we start counting as the first bus leaves for the day then it will not see another bus until it is half way along it's route.
Laura and Harriet from The Mount School explained it as follows:
Bus A and B set out at the same time, then 10 minutes later buses C and D set out. At this point none of the buses have advanced enough to meet another. Then, after 20 minutes, two more buses, E and F set out, and A and B meet. After 25 minutes A will meet D. 30 minutes into the cycle, A will meet F. G and H will set off. After 35 minutes A will meet H. Then at 40 minutes, the cycle will be complete, and A will meet one final bus, J, which is setting out. If bus A had been the first bus to set off, it would meet 5 other buses. If bus A had set off while the cycle was taking place, it would meet 8 other buses.
Well done to all of you who managed to crack this problem. | 798 | 3,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2016-44 | longest | en | 0.960647 |
http://www.jiskha.com/display.cgi?id=1203478163 | 1,498,573,091,000,000,000 | text/html | crawl-data/CC-MAIN-2017-26/segments/1498128321426.45/warc/CC-MAIN-20170627134151-20170627154151-00485.warc.gz | 565,707,002 | 4,335 | # Math: Calculus - Vectors
posted by .
A boat travels at a speed of 5 m/s in still water. The boat moves directly across a river that is 70m wide. The water in the river flows at a speed of 2 m/s. How long does it take the boat to cross the river? In what direction is the boat headed when it starts the crossing.
Textbook Answer: 15.3 m/s, 66.4 degrees to shore
My answer: 14 seconds and 22 degrees from the shore.
This is what I did: I found the speed of the resultant by doing the Pythagorean (5.4 m/s),then found the distance of the resultant by "5/70 = 2/x (28m)" and then using the Pythagorean (75.4m), and then I found the time by doing "t = d/s". Then, I found the direction of the boat by using trig, "tanx = 2/5"
• Math: Calculus - Vectors -
How can the textbook answer be 15.3 m/s if they are asking for the crossing time?
To go directly across, the boat must aim upstream so that the velocity component upstream relative to the water is 2 m/s. That makes its velocity component across the water sqrt[5^2 - 2^2) = sqrt 21 = 4.58 m/s. The crossing time is
70 m/4.58 m/s = 15.3 s. The pointing angle of the boat is cos^-1 4.58/5 = 23.7 degrees relative to the cross-stream direction, or 66.3 degrees relative to the shore.
• Math: Calculus - Vectors -
If the boat moves directly across then there is enough of the boats speed vector to counteract the river flow. That is 2m/s.
So, the vector parallel to the shore (the river flow) speed is 2m/s. The TOTAL speed of the boat is 5m/s. Now solve this for the speed perpendicular to the shore.
SQRT(5^2 - 2^2).
Divide that into the distance (70m).
Use any of the trig values to find the angle. | 471 | 1,658 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2017-26 | latest | en | 0.92295 |
https://www.r-bloggers.com/project-euler-problem-12-2/ | 1,575,713,239,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540497022.38/warc/CC-MAIN-20191207082632-20191207110632-00330.warc.gz | 840,122,444 | 51,722 | # Project Euler — problem 12
July 7, 2012
By
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.
Going to supper in 20 minutes. I’d like to type down my solution to the 12th Euler problem, just make my time count.
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, … We can see that 28 is the first triangle number to have over five divisors. What is the value of the first triangle number to have over five hundred divisors?
Here, two issues to be solved. One is to calculate the triangle number, which is simple as we known that nth triangle number, i.e. 1 + 2 + 3 + … + n = n * (n+1) / 2. The other issue is to find out the divisors of one given number, which is harder. The most straightforward way, also the most brute-forced, is to perform division calculations with all the numbers smaller than the given number N. In R, one could use `sum((N/1:N) == 0)` to get the sum of divisors. Well, it seems we get the solution. Use the first equation to calculate the first triangle number, and check whether the sum of divisors more than 500; if not (definitely not, the first is 1), move on to the next triangle number till you find the result.
However, this is so slow that I could’t take it. Using so-called integer factorization would speed up a lot. Once the list of prime factors is in hand, e.g. (n1 of p(rime)1, n2 of p2, …), the sum of divisors is easy to get, which is (n1 + 1) * (n2 + 1)*…*(nlast1 + 1).
?View Code RSPLUS
```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 ``` ```# helper funtion for factorization PrimeFactor <- function(x, prime = prime) { m <- length(prime) fac.count <- numeric(m) names(fac.count) <- prime # actually, a primality check could insert here for (i in 1:m) { prime.num <- prime[i] while (x %% prime.num == 0) { fac.count[i] <- fac.count[i] + 1 x = x / prime.num } while (x == 1) break } return(fac.count) } prime <- c(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97) # generate triangle numbers and count prime factors i <- 1 div.count <- 0 while (div.count <= 500) { triangle <- i * (i + 1) / 2 fac <- prime.fac(triangle, prime) div.count <- prod(fac + 1) i <- i + 1 } cat("The result is", i-1, "th triangle number:", triangle, "\n")```
Actually, I take some risks here, only using the prime number less than 100, which turned out OK. As it is to find the first triangle number with more than 500 divisors, i.e. the smallest one, the composite number with smaller primes should have bigger chance. And by checking the remainder, the risk of mis-calculation could also be avoided.
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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't. | 959 | 3,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-51 | latest | en | 0.874504 |
http://mathhelpforum.com/algebra/68389-expressing-terms-b-print.html | 1,513,326,744,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948567785.59/warc/CC-MAIN-20171215075536-20171215095536-00321.warc.gz | 193,273,731 | 3,090 | # Expressing In Terms of a And b
• Jan 15th 2009, 02:27 PM
Shapeshift
Expressing In Terms of a And b
If log (36) = a and log (125) = b
find log (1/12) in terms of a and b
I tried to do things like applying the product formula to log (36), making it log (12) + log (3), and then turning log (12) into -log (1/12), but could not get any further. Any help is appreciated.
• Jan 15th 2009, 02:38 PM
hmmmm
you would set k(log36/log125)=log(1/12) find k and the rewrite the log36 and log125 with a and b (im assuming log is the natural log in this case ln??)
• Jan 15th 2009, 04:16 PM
Shapeshift
I got...
log (1/12) = -a/2+b/3-1
could anyone confirm this? thanks!
• Jan 15th 2009, 08:14 PM
Soroban
Hello, Shapeshift!
Assuming that the logs are base-ten, you are right!
Quote:
If $\log(36) \,=\, a\:\text{ and }\:\log(125) \,=\, b$
find $\log\left(\tfrac{1}{12}\right)$ in terms of $a\text{ and }b.$
We have: . $\begin{array}{cccccccc}
\log(36) \:=\:a & \Rightarrow & \log(6^2) \:=\:a & \Rightarrow & 2\log(6) \:=\:a & \Rightarrow & \log(6) \:=\:\frac{a}{2} \\ \\[-4mm]
\log(125)\:=\:b & \Rightarrow & \log(5^3) \:=\:b & \Rightarrow & 3\log(5) \:=\:b & \Rightarrow & \log(5) \:=\:\frac{b}{3}
\end{array}$
Therefore: . $\log\left(\frac{1}{12}\right) \;=\;\log\left(\frac{5}{6\cdot10}\right) \;=\;\log(5) - \log(6) - \log(10) \;=\;\boxed{\frac{b}{3} - \frac{a}{2} - 1}$ | 546 | 1,371 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.15625 | 4 | CC-MAIN-2017-51 | longest | en | 0.6344 |
https://www.victoriaadvocate.com/news/2011/apr/19/ds_money_matters_042011_136649/ | 1,516,511,625,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084890187.52/warc/CC-MAIN-20180121040927-20180121060927-00652.warc.gz | 1,001,207,277 | 18,888 | # Dave Sather's Money Matters: What does 'devaluing the dollar' mean?
April 19, 2011 at 4:02 p.m.
Updated April 18, 2011 at 11:19 p.m.
Many times, financial news outlets reference the "devaluation of the dollar." What does this mean and what are the implications?
Recently, Carol and I were in Spain. She was speaking at an academic conference. I was there merely to carry her bags.
One particular day, as we strolled through town, my bride started to get hungry. With our command of the Spanish language being rather thin, we decided to venture into the golden arches of McDonald's.
Although it is rather ironic that we would travel halfway around the world to eat at Mickey D's, it was familiar, and we could order simply by pointing and not butchering the language any further.
Our order was easy enough: a Big Mac, a chicken sandwich, two medium fries and two medium drinks. The total bill came to 13 euros. As I ate my Big Mac, I started doing some math - converting euros into dollars. Holy crap! I just paid \$21 United States for some greasy fries and grade D beef. Haven't these people ever heard of a "value meal?"
Could it really be that much more expensive to produce a fast-food meal in Spain than in the United States? Maybe that is part of it. But as I got more and more irritated at the price of my meal, I pulled out my computer (thankfully McDonald's had Wi-Fi) and started researching the correlation between the dollar and the euro.
There, plain as day, was a chart showing me that just in the last nine months the dollar had lost 18 percent of its purchasing power relative to the euro. I started to get a sinking feeling that this trip was going to cost quite a bit more than I had anticipated.
The dollar's relationship to the euro is not unique. Over the same timeframe the U.S. dollar lost 21 percent of its value relative to the Australian dollar and 23 percent relative to the Swiss franc.
Although the McDonald's experience is just one example, it gives you an idea of what happens when our currency is devalued.
In general, when you devalue one currency, things manufactured in other countries that use other currencies become more expensive - all things being equal.
When we travel to other countries, our money buys less in terms of food, lodging and virtually any other service.
Furthermore, it makes imports into the United States more expensive for Americans. This starts to give you an idea of why this game is played.
The U.S. government will intentionally devalue the dollar knowing that foreign companies will have a harder time selling their goods to Americans. Additionally, when the U.S. dollar is devalued, it makes the products manufactured in the United States more affordable in foreign countries. This can stimulate U.S. exports to other nations.
On the surface, this may seem like a great way to prop up a country in need of additional sales and jobs. However, we are not the only ones playing this game. Any country with its own currency is madly running the printing presses day and night - all in an effort to make its products sell better.
Unfortunately, we do not live on a self-contained island. We will continue to import things - such as oil. As such, it is the average consumers who get hurt as their paychecks simply don't buy as much of the goods and services they need each day.
Whether you chalk it up to currency devaluation or inflation, the outcome is the same. A dollar just doesn't buy what it used to.
Dave Sather is a Victoria Certified Financial Planner and owner of Sather Financial Group. His column, Money Matters, publishes every other Wednesday. | 780 | 3,636 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2018-05 | latest | en | 0.98094 |
https://www.coursehero.com/file/11109572/Final-Exam-Review-Sheet/ | 1,632,641,092,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057830.70/warc/CC-MAIN-20210926053229-20210926083229-00052.warc.gz | 729,643,756 | 98,111 | # Final Exam Review Sheet - 18.01 Calculus Final Exam at...
• Test Prep
• Zachchen
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18.01 Calculus Jason Starr Final Exam at 9:00am sharp Fall 2005 Tuesday, December 20, 2005 More 18.01 Final Practice Problems Here are some further practice problems with solutions for the 18.01 Final Exam. Many of these problems are more difficult than problems on the exam. Goal 1. Differentiation. x 1.1 Find the equation of every tangent line to the curve y = e containing the point ( 1 , 0). This is not a point on the curve. 1.2 Let a and b be positive real numbers. Find the equation of every tangent line to the ellipse with implicit equation, 2 2 x y + = 1 , a 2 b 2 containing the point (2 a, 2 b ). This is not a point on the ellipse. 1.3 Let a be a real number different from 0. Use the definition of the derivative as a limit of difference quotients to find the derivative to the following function, 1 f ( x ) = , x at the point x = a . 1.4 Use the definition of the derivative as a limit of difference quotients to find the derivative of the following function, f ( x ) = tan( x ) , at the point x = 0. You may use without proof that the following limits exist and have the given values, sin ( x ) 1 cos( x ) lim = 1 , lim = 0 . x 0 x x 0 x 1.5 For x > 0, let f ( x ) be the function, x f ( x ) = e . Thus the inverse function, y = f 1 ( x ) , 1
18.01 Calculus Jason Starr Final Exam at 9:00am sharp Fall 2005 Tuesday, December 20, 2005 satisfies the equations, e y = x, and y = ln( x ) . Compute the derivative, dy . dx Goal 2. Sketching graphs. 2.1 Sketch the graph of the function, 1 2 1 f ( x ) = + . x 1 x x + 1 2.2 Sketch the implicit function, y 2 xy x 2 = 1 . 2.3 Sketch the graph of the function, x 2 x 2 f ( x ) = + . x + 1 x 1 Goal 3. Applications of differentiation. 3.1 A sculpture has the form of a right triangle. The material used for the vertical leg has twice the cost of the material used for the horizontal leg. The length of the hypotenuse is fixed (thus its cost is irrelevant). What ratio of vertical leg to horizontal leg minimizes the total cost of the material? 3.2 A farmer has a fence running diagonally across her property at a 45 degree angle to the north- south and east-west lines. She decides to build a corral by adding a length b a of fence running north-south, a length b a of fence running east-west, and then connect the two corners with 2 length b of fence running north-south and east-west. Thus, the total new length of fence needed is 4 b 2 a , and the corral has the form of a square of length b , with a small isosceles triangle of leg length a removed from one corner (where the square corral meets the pre-existing diagonal fence). What ratio of a to b gives maximal area of the corral for a fixed length of new fence? 3.3 An icicle has the shape of a right circular cone whose ratio of length to base radius is 10. Assuming the icicle melts at a rate of 1 cubic centimeter per hour, how fast is the length of the icicle decreasing when it is 10 centimeters long? | 852 | 3,059 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2021-39 | latest | en | 0.852374 |
http://www.binghamton.edu/mpr/ask-a-scientist/entry.html?id=279 | 1,462,362,622,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860123023.37/warc/CC-MAIN-20160428161523-00191-ip-10-239-7-51.ec2.internal.warc.gz | 379,903,687 | 12,123 | ### When my mom makes pasta, she puts salt in the boiling water. After a few minutes, the water bubbles. Why is this?
School: Seton Catholic at All Saints School
Teacher: Mr. Martinkovic
Career Interest: Sports player or work at my dad’s office
### Answer from Douglas W. Green, EdD
Former principal at Woodrow Wilson Elementary in Binghamton, NY
Research area: Leadership, Learning Theory & Social Media
Interests/hobbies: Playing banjo, biking, golfing and reading
Family: Daughter Lena, age 27, who is an animator for Nickelodeon in New York City
Check out Doug's blog, a site for educators who don't have as much time to read as he does http://www.drdouggreen.com.
All food contains some salt. If you heat food containing salt in water without salt, some of the salt in the food will migrate out of the food and into the water. If too much salt leaves the food it can taste bland. When you add salt to water, the temperature at which the water boils will go up so the food will cook a bit faster. Mom may not realize this, but this explains why many people add salt to pasta water.
As for boiling, it is important to understand that water molecules move faster as you heat them. This is true for all materials. If water molecules move fast enough, they can go from being a liquid, where the molecules are close together, to being a gas where the molecules are far apart. In the case of water, this happens when the temperature reaches about 212 degrees Fahrenheit or 100 degrees Celsius. Temperature is just a measure of the average speed to the molecules in a sample.
Water molecules are a bit like little magnets, which is why they are attracted to each other. If they get cold enough, they lock in to a crystal formation as solid ice. If you heat ice, the molecules vibrate until the are no longer “frozen” in place and become liquid water. When the molecules move fast enough, they can leave the surface and become water vapor, which is a gas like the air you breath. The air around you always contains some water vapor in addition to oxygen, nitrogen, small amounts of other chemicals, and dust. This is a good thing as water in the air is necessary for good health.
Below the water’s surface in your pasta pan, the rough edges of a pasta pan that are too small to see help gas bubbles form. Since these bubbles are much less dense than the surrounding water, they rise to the surface and become part of the air in the room. As a pan of water boils, water molecules gradually leave. If you don’t pay attention to your boiling pasta, all of the water can boil into the room. This is dangerous and smelly as once the water is all gone, the contents will get much hotter and may catch fire. Water is so important to life on Earth that understanding it is one of the most important things that scientists and citizens can do.
Dr. Doug Green blogs at DrDougGreen.Com for educators and parents who don’t have as much time to read and surf as he does.
Last Updated: 9/18/13 | 659 | 2,984 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2016-18 | longest | en | 0.948209 |
https://electronics2electrical.com/11553/ | 1,563,610,095,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526489.6/warc/CC-MAIN-20190720070937-20190720092937-00501.warc.gz | 388,111,348 | 22,344 | # Torque developed by a three-phase, 400V, induction motor is 100 N -m. If the applied voltage is reduced to
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Torque developed by a three-phase, 400V, induction motor is 100 N -m. If the applied voltage is reduced to 200-V, the d'veloped torque will be
Torque developed by a three-phase, 400V, induction motor is 100 N -m. If the applied voltage is reduced to 200-V, the d'veloped torque will be 25 N -m
## Related questions
The torque developed by any 3-phase induction motor at 0.8 p.u. slip is (A) Full-load torque (B) unstable torque (C) starting torque (D) break down
The speed of a three-phase induction motor is controlled by variable voltage variable frequency control (i.e. keeping V/f constant). As the frequency is reduced, the slip at maximum torque (1) Decreases (2) Increases (3) Remains constant (4) None of the above
A 400 V, 50 Hz three-phase induction motor rotates at 1440 rpm on full -load. The motor is wound for
In a double squirrel cage induction motor, good starting torque is produced as (A) Large current flows through the outer high resistance winding. (B) Large current flows through the inner high resistance winding. (C) Small current flows through the outer high resistance winding. (D) Large current flows through the inner low resistance winding.
In an induction motor if the flux density is reduced to one-half, of its normal value then the torque will (1) reduce to one-half (2) reduce to one-fourth (3) remain unchanged (4) increase four times
In a single-phase induction motor when the rotor is stationary and voltage is applied to the stator, then (A) the flux is constant (B) the flux first decreases and then increases in the same direction (C) the flux increases and decreases in the opposite direction with the same magnitude (D) current flows through the rotor
In a 3-phase slip-ring induction motor, the slip at some speed is 0.05. The speed is reduced by inserting an external resistance in the rotor circuit and the slip is 0.15. If the rotor winding resistance per phase is 0.2 Ω, the external resistance is (A) 0.2 Ω (B) 0.3 Ω (C) 0.4 Ω (D) 0.5 Ω
What is mean by plugging in three phase induction motor?
An 8-pole, 50Hz, three-phase induction motor is running at 705rpm and has a rotor copper loss of 5kW. Its rotor input is (A) 5.06 kW (B) 0.3 kW (C) 100 kW (D) 83.33 kW
How is a three phase cage induction motor self starting?
The slip of 400 V, three phase, 4 -pole induction motor when rotating at 1440 rpm is
A 3-phase, 4-pole, 50Hz induction motor runs at a speed of 1440 rpm. The rotating field produced by the rotor rotates at a speed of ________ rpm with respect to stator field. (A) 1500 (B) 0 (C) 1440 (D) 60
speed control methods of three phase induction motor?
A change of 4% in the supply voltage to an induction motor will produce a change of approximately
A capacitor-start single phase induction motor is switched on to supply with its capacitor replaced by an inductor of equivalent reactance value. It will?
In a single-phase induction motor, the operating direction is determined by the ________ A) Rotor B) Stator C) Starting circuit D) None of these
In a split-phase induction motor, the resistance/reactance ratios of windings are such that (A) current in the auxiliary winding leads the current in the main winding (B) current in the auxiliary winding lags the current in the main winding (C) both windings develop the same starting torque (D) both windings develop high starting current
In a 2 kW, 200 V, 1000 rpm. DC series motor the torque at full load was found to be 0.3 N -m. The torque at half full load in N -m is :
how can we reverse the direction of rotation of three phase induction motor ?
How does a single phase induction motor work?
explain the working of single phase induction motor.
A balanced three phase star connected load is supplied from a three phase 400V 50 Hz supply. The resistance for per phase is 10 Ohm. Find the value of phase current, line current, power factor and total power consumed.
Split phase induction motor
Single Phase Induction Motor
What is the slip (as a percentage) when a 6 pole single phase induction motor operating on 220 V and 60 Hz is rotating at speed of 1140 rpm? A) 60% B) 5% C) 10% D) None of these
What is the definition of induction motor?
What is the function of induction motor?
At which time power factor of the induction motor is low?
The number of poles in a phase induction motor at 50 Hz and running at 1500 rpm will be? | 1,130 | 4,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2019-30 | longest | en | 0.912052 |
https://www.doorsteptutor.com/Exams/CBSE/Class-12/Economics/Questions/With-Solutions-Part-1.php | 1,542,075,868,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741192.34/warc/CC-MAIN-20181113020909-20181113042909-00427.warc.gz | 853,133,193 | 16,305 | # CBSE (Central Board of Secondary Education- Board Exam) Class-12 Economics: Questions 1 of 523
» Consumer's Equilibrium and Demand » Demand » Elasticity of Demand
Essay Question▾
### Describe in Detail
Explain the geometric method of measuring price elasticity of demand.
## Explanation
In geometric method we measure elasticity of demand at different points on demand curve. In this method elasticity of demand at any point on straight line demand curve is ratio of lower segment and upper segment of demand curve.
It is represented as:
=
In this figure Point A represent = =0
Point B represent = = infinite (∞)
Point D represent = = 1
Point E represent = > 1
Point F represent = < 1 | 155 | 699 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2018-47 | latest | en | 0.889516 |
http://www.apmonitor.com/me575/index.php/Main/LinearMultivariateRegression?action=sourceblock&num=2 | 1,712,986,637,000,000,000 | text/plain | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00060.warc.gz | 33,231,925 | 1,612 | import numpy as np from scipy.stats import linregress import statsmodels.api as sm import matplotlib.pyplot as plt from gekko import GEKKO # Data x = np.array([4,5,2,3,-1,1,6,7]) y = np.array([0.3,0.8,-0.05,0.1,-0.8,-0.5,0.5,0.65]) # calculate R^2 def rsq(y1,y2): yresid= y1 - y2 SSresid = np.sum(yresid**2) SStotal = len(y1) * np.var(y1) r2 = 1 - SSresid/SStotal return r2 # Method 1: scipy linregress slope,intercept,r,p_value,std_err = linregress(x,y) a = [slope,intercept] print('R^2 linregress = '+str(r**2)) # Method 2: numpy polyfit (1=linear) a = np.polyfit(x,y,1); print(a) yfit = np.polyval(a,x) print('R^2 polyfit = '+str(rsq(y,yfit))) # Method 3: numpy linalg solution # y = X a # X^T y = X^T X a X = np.vstack((x,np.ones(len(x)))).T # matrix operations XX = np.dot(X.T,X) XTy = np.dot(X.T,y) a = np.linalg.solve(XX,XTy) # same solution with lstsq a = np.linalg.lstsq(X,y,rcond=None)[0] yfit = a[0]*x+a[1]; print(a) print('R^2 matrix = '+str(rsq(y,yfit))) # Method 4: statsmodels ordinary least squares X = sm.add_constant(x,prepend=False) model = sm.OLS(y,X).fit() yfit = model.predict(X) a = model.params print(model.summary()) # Method 5: Gekko for constrained regression m = GEKKO(remote=False); m.options.IMODE=2 c = m.Array(m.FV,2); c[0].STATUS=1; c[1].STATUS=1 c[1].lower=-0.5 xd = m.Param(x); yd = m.Param(y); yp = m.Var() m.Equation(yp==c[0]*xd+c[1]) m.Minimize((yd-yp)**2) m.solve(disp=False) c = [c[0].value[0],c[1].value[1]] print(c) # plot data and regressed line plt.plot(x,y,'ko',label='data') xp = np.linspace(-2,8,100) slope = str(np.round(a[0],2)) intercept = str(np.round(a[1],2)) eqn = 'LstSQ: y='+slope+'x'+intercept plt.plot(xp,a[0]*xp+a[1],'r-',label=eqn) slope = str(np.round(c[0],2)) intercept = str(np.round(c[1],2)) eqn = 'Constraint: y='+slope+'x'+intercept plt.plot(xp,c[0]*xp+c[1],'b--',label=eqn) plt.grid() plt.legend() plt.show() | 700 | 1,874 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.84375 | 3 | CC-MAIN-2024-18 | latest | en | 0.35788 |
https://www.daniweb.com/programming/software-development/threads/107791/help-starting-a-program-about-a-circle | 1,547,753,679,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659063.33/warc/CC-MAIN-20190117184304-20190117210304-00487.warc.gz | 751,380,878 | 13,985 | ill show the problem in the book first then ill tell you what i know.
___________________________________________________________________________
Then following formula gives the distance between two points (x1,y1) and (x2,y2) in the Cartesian plane: √ (x2-x1)^2 + (y2-y1)^2 its the distance formula..hard to type on here
Given the center and a point on the circle, you can use this formula to find the radius of the circle. Write a program that prompts the user to enter the center and a point on the circle. The program should then output the circles radius, diameter, circumference, and area. The program must have at least the following functions:
distance - this function takes as its parameters four numbers that represent two points in the plane and returns the distance between them.
radius - this function takes as its parameters four numbers that represent the center and a point on the circle, calls the function, distance, to find the radius of the circle, and returns the circles radius.
circumference: this function takes as its parameter a number that represents the radius of the circle and returns the circle's circumference (if r is the radius, the circumference is 2*pie*r)
area: this function takes as its parameter a number that represents the radius of the circle and returns the circles area. (if r is the radius, the area is (pie)(r)^2 ...thats pie times r squared.
assume that pie=3.1416
____________________________________________________________________________
I'm just confused on how to set this all up. so i just typed in all the code i have an idea about and maybe one of you guys can help set this thing up for me and give your thoughts on how to do this.. Any information is appreciated!!! Thanks guys.
``````//I have to prompt to enter a center and a point on the circle so maybe something like this:
cout<<"Enter a center point: ";
cin>>x1, y1;
cout<<"Enter a point on the circle: ";
cin>>x2, y2;
//my functions will be something like this: maybe im not sure
int distance (int x1, int y1, int x2, int y2)
{
int dx = x2 - x1;
int dy = y2 - y1;
float dsquared = dx*dx + dy*dy;
float result = sqrt (dsquared);
cout<<"The distance of the circle is "<<result<<"."<<endl;
}
int radius (int x1, int y1, int x2, int y2)
{
float radius = distance (x1, y1, x2, y2);
cout<<"The radius of the circle is "<<result<<"."<<endl;
}
{
float circumference = 3.1416 * (radius * 2);
cout<<"The circumference of the circle is " <<circumference<< "."<<endl;
//ok im not sure on this one because he says the function "area" should be made void and
//the return value should come from a reference parameter.
{
return area;
}``````
This may not be your complete code. As far as I can see, you may need
- function prototypes before calling them
- to be careful with types
Furthermore since c++ is there for oop, change whole thing into oop approach (i.e. to a circle class)
our professor wanted to challenge us to not declare the circle class or believe me haha, i definately would. yea it wasn't all of my code. I was having a hard time starting it, but believe it or not, I looked at what i wrote and started a very very rough draft from it. Here is what the compiler says first, then ill put my whole code up:
I think the problems i have in my mind are the declarations on what to have right beside the name in the functions... for example int circumference().. the other problems im having ill say something above the code... thanks once again!!!!!
line 8: Error: Multiple declaration for y1.
line 71: Error: Too many arguments in call to "area()".
line 71: Error: A value of type void is not allowed.
line 84: Error: "area(float)" cannot return a value.
line 89: Error: The operand "*diameter" cannot be assigned to.
line 90: Error: Cannot return int(*)(float) from a function that should return int.
6 Error(s) detected.
``````#include <iostream>
using namespace std;
//declare variables (I'm not really sure why it says I declared y1 twice)
int x1;
int x2;
int y1;
int y2;
//Declare functions
int distance();
int circumference();
void area();
int main()
{
//Get a center point
cout<<"Enter a center point: ";
cin>>x1, y1;
//Get another point on the circle
cout<<"Enter a point on the circle: ";
cin>>x2, y2;
//Call the functions
int distance();
void area();
int circumference();
int diameter();
//distance return
cout<<"The distance of the circle is: ";
int distance();
//area return
cout<<"The area of the circle is: ";
void area();
cout<<"The radius of the circle is: ";
//circumference return
cout<<"The circumference of the circle is: ";
int circumference();
//diamter
cout<<"The diameter of the circle is: ";
int diameter();
}
int distance (int x1, int y1, int x2, int y2)
{
int dx = x2 - x1;
int dy = y2 - y1;
float dsquared = dx*dx + dy*dy;
float result = sqrt (dsquared);
return result;
}
int radius (int x1, int y1, int x2, int y2)
{
float radius = distance (x1, y1, x2, y2);
return result;
}
//The (float radius) on this one and the void area one I'm not sure about. The parameters
//confuse me, sorry, i'm a newbie. haha
{
float circumference = 3.1416 * (radius * 2);
return circumference;
}
//this one im having a hard time because our professor said the function area should be
//made void and the return value should come from a reference parameter(which is before
//main I think, correct me if im wrong please)
{
return area;
}
//see area, circumference for confusion
{
return diameter;
}``````
take a look at the [itex] library
pow can be used to square the numbers.
Define Pi as a global , then just use the variable.
The area function returnd a float, so it's NOT void,but :
``float area(float radius)``
You're calling your functions wrongly. They should look more like this.
``rad = radius(x1, y1, x2, y2);``
rad, is the value you'd print out as well as pass to your other functions like
``area = areaOfCir( rad );`` | 1,478 | 5,900 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-04 | latest | en | 0.863888 |
https://subscanner.com/3_RhISgoXUs | 1,653,775,348,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663021405.92/warc/CC-MAIN-20220528220030-20220529010030-00468.warc.gz | 610,152,307 | 4,912 | # Every Force in Nature (Theory of Everything, Part III)
We know the universe has stuff in it called "matter" that takes up space.
And the universe isn't just static - things happen, too!
To be more precise, matter interacts with other matter, like the pull of the earth and
the moon on each other, or the mutual repulsion of your feet and the ground.
We might think that these interactions are just traffic laws imposed by an obtuse pan-galactic
regulatory committee… but in fact, the real answer is much deeper, and more bizarre: interactions
happen simply because you (and by you I mean "the universe") can measure things differently
in different places.
But wait a second: regardless of whether I measure a person to be 6 feet or 183 cm tall,
they're still the same height!
So measurement, by itself, is meaningless, but as surprising as it sounds, that meaninglessness
is exactly what causes the fundamental forces of nature.
Now, often on MinutePhysics I use analogies to help simplify complicated physical phenomena
while still getting the essence of the point across.
But today we're going to witness a real live application of the theory behind the standard
model, and while for simplicity's sake we're going to ignore quantum effects and apply
it to economics instead, this is literally the same math as the standard model.
Suppose, for example, that I gave you \$2 for a \$2 sheep – nothing has really happened,
or no value has been transferred, because I could sell you the sheep back and end up
with my same \$2 again.
Ignoring the fact that maybe I like sheep more than money, which is why we'd trade in
the first place, the real value as measured by our "value-ometer", which we call money,
is unchanged.
What's more, the amount of money we pay for things is really an arbitrary scale, which
is obvious when you remember that different countries have different currencies.
But of course countries can't interact financially without a way of converting the way they measure
value.
So, now suppose I want to buy a sheep in Canada – there are two things that'll affect the
price: first, Canadians may just value sheep less than I do, and second, their dollar may
be different from my US dollar… so perhaps the sheep will only cost \$3 Canadian.
Which of course means I can go to Canada, convert my 2 US dollars to 4 Canadian dollars,
spend three of them on the sheep, and come back to the US with my sheep, plus one canadian
dollar (which converts to fifty US cents).
If I then sell the sheep back to someone in the US for the \$2 it's worth, I'll now have
This is very different from the case of buying and selling a sheep inside the US, where no
real value changed hands.
Here, just by making a "currency exchange" part of my sheepish journey, I was able to
transfer real value from Canada to the US, and this transfer of value happened solely
because we measure value differently in different places.
It's like stealing, only legal!
Now, don't all go running off to buy Canadian sheep and sell them in the US… in the real
world shepherds (and money exchangers) realize that they're losing out, and the price of
sheep (and CAD) will adjust to take this into account and minimize all transfers of real
value, or "stealing".
But "making money from nothing" does happen if you can act before the market adjusts – it's
called "Arbitrage," and anytime it's possible, it means that the economy isn't in an equilibrium,
or optimal, state.
So much for the "invisible-hand" of the market…
But anyway, in physics, this effect of stealing real value is called "momentum transfer"…
or in day-to-day terms, a force.
And now we'll see why!
Suppose, instead of one border and one exchange rate, we have a whole row of countries that
can each exchange money with their neighbors:
Now if I want to pull off my arbitrage shenanigans by selling a sheep in Iran, I'll have to transfer
the money back to the US, which in this case means it'll be exchanged at every border along
the way.
But this series of measurement conversions looks like it's "moving"… like, it's an
"exchange-rate-particle" that gets created in Iran, carries value from Iran to the US,
then disappears!
Wait, let's see that again: buy a sheep, bring it to Iran, sell it, money changes from rial
to rupee to mirian to rupee to pound to dollar to dollar… and I end up with more money
than I bought the sheep for to begin with!
So that's it: real standard model physics in your own barnyard.
And hopefully now you can see why measuring things differently in different places inevitably
gives rise to a long-range interaction mediated by a particle!
For example, the electromagnetic potential tells us how electric charge is measured differently
at different places… it's the "electron-exchange-rate", and the excitations in the electromagnetic
field are particles, which we call photons.
Instead of transferring monetary value, these photons transfer momentum from one electron
to another, and if you add up a whole bunch of these momentum transfers, you get something
that we call a force!
But isn't a photon a particle of light and not a "force"!?!
Well, once you have the idea of an "exchange-particle", you don't actually need the electrons at the
endpoints anymore… you could just have that particle moving through empty space on its
own.
That's why photons are both the particles that mediate the electromagnetic force, and
bonafide particles on their own!!
In fact, all the forces we know, like the electromagnetic interaction, the strong interaction,
weak interaction, and gravitational interaction, work this same way at a fundamental level:
an "exchange-particle," which physicists crazily call a "gauge boson", transfers momentum and
energy between two matter particles.
And this is what Newton was trying to get at when he said "for every action there is
an equal and opposite reaction"… we'll cut him some slack since he lived in the seventeenth
century, but what he really should have said was "for every interaction you need an exchange
particle."
And maybe if he had known this, he could have turned sheep into gold. | 1,359 | 6,094 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.734375 | 3 | CC-MAIN-2022-21 | latest | en | 0.964111 |
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